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Hypothesis Testing Calculator Online

Hypothesis testing is a foundational method used in statistics to infer the validity of a hypothesis about a population parameter. The Hypothesis Testing Calculator facilitates this process by automating the computations necessary for the t-test , a method used to compare sample means against a hypothesized mean or against each other. Let’s delve into the formulas this calculator uses to execute one-sample and two-sample t-tests.

One-Sample t-Test

Two-sample t-test, equal variances:, unequal variances (welch’s t-test):, table of critical t-values.

These values are crucial in hypothesis testing as they help define the threshold for significance, assisting users of the calculator in interpreting their results accurately.

The calculated t-value is 3.00. Using the critical t-values table, at 95% confidence level and 35 degrees of freedom, the critical value is approximately 2.030. Since 3.00 > 2.030, the null hypothesis is reject, indicating a significant difference from the hypothesize mean.

Most Common FAQs

Yes, while the t-test is specifically design for means, the principles of hypothesis testing apply to other parameters such as proportions and variances. Which can also be tested using appropriate versions of hypothesis tests such as the z-test and F-test.

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$$ \dfrac{\overline{x} - μ_0}{\dfrac{σ}{\sqrt{n}}} $$

$$ \dfrac{\overline{x} - \overline{y}}{\sqrt{\dfrac{σ^2_x}{n_1} + \dfrac{σ^2_y}{n_2}}} $$

$$ \dfrac{\stackrel{\text{^}}{p} - \ p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}} $$

$$ \dfrac{\stackrel{\text{^}}{p_1} - \stackrel{\text{^}}{p_2}}{\sqrt{\stackrel{\text{^}}{p}(1-\stackrel{\text{^}}{p})(\dfrac{1}{n_1} + \dfrac{1}{n_2})}} $$

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Try the test statistic calculator to estimate the t-value of a given dataset within seconds by providing a couple of simple inputs.  This calculator takes sample mean, population mean, standard deviation, and sample size into account to calculate t statistics precisely. 

What Is T Statistic And Student's T-Test?

A particular statistical calculation that figures out the relationship between the sample and its population is known as the test statistics.  It's a score that is used in the hypothesis test and informs about how likely the results are under the assumption that the null hypothesis is true. Moreover, a student’s t-test is used to evaluate hypotheses about the population mean . It tells whether you should support or reject the null hypotheses.

Test Statistic Formula?

The test statistic formula depends on the population standard deviation (σ) whether it's known or unknown. Let's take a look at the following general test statistic equations:

For UnKnown Standard Deviation: t = \(\frac{\overline{x} - μ_0}{\frac{s}{\sqrt{n}}}\)

For Known Standard Deviation: t = \(\frac{\overline{x} - μ_0}{\frac{σ}{\sqrt{n}}}\)

• \( \overline{x} \) indicates the sample mean
• μ shows the population mean
• n is the representative of the sample size
• s is the standard deviation

Use the above mentioned formula sourced from Wikipedia to compute your test statistic.

How To Calculate Test Statistic?

There are different standard cases that you must know to perform test statistics. Below we have four standard cases for which t value formulas differ.

One Population Mean:

Calculating one population mean, if the variable is numerical and one population or one group is being studied.  For instance, you believe that the students of O level spend $20 a day, here the variable is numerical and the population is the students of O level. In this case, it's good to use one population mean. 

Test Statistic for One Population Mean = \(\frac{\overline{x} - μ_0}{\frac{σ}{\sqrt{n}}}\)

Comparing Two Means:

This test is good when the variable is numerical and you have to compare two populations or groups. For this, you must choose two separate random samples from each population.

Test Statistic Comparing Two Means = \(\frac{\overline{x} - \overline{y}}{\sqrt{\frac{σ^2_x}{n_1} + \frac{σ^2_y}{n_2}}}\)

Single Population Proportion:

Perform this calculation, when the variable is categorical such as genders, workers/ unemployed and one population has to be considered. 

Test Statistic for a Single Population Proportion =   \(frac{\stackrel{\text{^}}{p} -\ p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\)

Two Population Proportions:

Use it when the variable is categorical, and you have to consider the proportion of two individuals having specific characteristics of two different categories such as male/female. 

Test Statistic for Two Population Proportions = \( \frac{\stackrel{\text{^}}{p_1}-\stackrel{\text{^}}{p_2}}{\sqrt{\stackrel{\text{^}}{p}(1-\stackrel{\text{^}}{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \)

Practical Example:

Let us make a supposition for a cricket series in which Jack has an average score of 66 or consecutive 16 matches. Now as you better know an average batting average for a player is 40 (maximum). Keeping in view the deviation in scoring that is 4 in the case, what are the performance stats of Jack? Now how to find the test statistics ?

Data Given:

\(\overline{x}=66\) \(n=16\) \(μ=40\) \(σ=4\)

Calculations:

\(\text{Test Statistic}=\frac{\overline{x} - μ_0}{\frac{σ}{\sqrt{n}}}\)

\(\text{Test Statistic}=\frac{66 - 40}{\frac{4}{\sqrt{16}}}\)

\(\text{Test Statistic}=\frac{26}{\frac{4}{4}}\)

\(\text{Test Statistic}=\frac{26}{1}\)

\(\text{Test Statistic}=26\)

Now as the computed value is 26, find a Standardized test statistic calculator to verify it. With it, you will have the correct results in fractions of seconds. 

Who Is The Father of Student’s T-Distribution?

Gosset was a talented statistician who proposed the theory of student’s t-distribution in the year 1908.

Is The Test Statistic The Z Value?

A z score is a value that shows a standard deviation above or below the mean. Meanwhile, it represents the relationship of value with the mean of a group. 

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Section 10.4: Hypothesis Tests for a Population Standard Deviation

• 10.1 The Language of Hypothesis Testing
• 10.2 Hypothesis Tests for a Population Proportion
• 10.3 Hypothesis Tests for a Population Mean
• 10.4 Hypothesis Tests for a Population Standard Deviation
• 10.5 Putting It Together: Which Method Do I Use?

By the end of this lesson, you will be able to...

• test hypotheses about a population standard deviation

For a quick overview of this section, watch this short video summary:

Before we begin this section, we need a quick refresher of the Χ 2 distribution.

The Chi-Square ( Χ 2 ) distribution

Reminder: "chi-square" is pronounced "kai" as in sky, not "chai" like the tea .

If a random sample size n is obtained from a normally distributed population with mean μ and standard deviation σ , then

has a chi-square distribution with n-1 degrees of freedom.

Properties of the Χ 2 distribution

• It is not symmetric.
• The shape depends on the degrees of freedom.
• As the number of degrees of freedom increases, the distribution becomes more symmetric.
• Χ 2 ≥0

Finding Probabilities Using StatCrunch

We again have some conditions that need to be true in order to perform the test 

• the sample was randomly selected, and
• the population from which the sample is drawn is normally distributed

Note that in the second requirement, the population must be normally distributed. The steps in performing the hypothesis test should be familiar by now.

Performing a Hypothesis Test Regarding σ

Step 1 : State the null and alternative hypotheses.

Step 2 : Decide on a level of significance, α .

Step 4 : Determine the P -value.

Step 5 : Reject the null hypothesis if the P -value is less than the level of significance, α.

Step 6 : State the conclusion.

In Example 2 , in Section 10.2, we assumed that the standard deviation for the resting heart rates of ECC students was 12 bpm. Later, in Example 2 in Section 10.3, we considered the actual sample data below.

( Click here to view the data in a format more easily copied.)

Based on this sample, is there enough evidence to say that the standard deviation of the resting heart rates for students in this class is different from 12 bpm?

Note: Be sure to check that the conditions for performing the hypothesis test are met.

[ reveal answer ]

From the earlier examples, we know that the resting heart rates could come from a normally distributed population and there are no outliers.

Step 1 : H 0 : σ = 12 H 1 : σ ≠ 12

Step 2 : α = 0.05

Step 4 : P -value = 2P( Χ 2 > 15.89) ≈ 0.2159

Step 5 : Since P -value > α , we do not reject H 0 .

Step 6 : There is not enough evidence at the 5% level of significance to support the claim that the standard deviation of the resting heart rates for students in this class is different from 12 bpm.

Hypothesis Testing Regarding σ Using StatCrunch

Let's look at Example 1 again, and try the hypothesis test with technology.

Using DDXL:

Using StatCrunch:

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Hypothesis testing for the mean Calculator

State the null and alternative hypothesis:

Calculate our test statistic z:.

Determine rejection region:

What is the answer, how does the hypothesis testing for the mean calculator work, what 1 formula is used for the hypothesis testing for the mean calculator, what 7 concepts are covered in the hypothesis testing for the mean calculator.

• alternative hypothesis
• hypothesis testing for the mean
• null hypothesis
• test statistic

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Two Population Calculator

Confidence interval.

Hypothesis Testing

When computing confidence intervals for two population means, we are interested in the difference between the population means ($ \mu_1 - \mu_2 $). A confidence interval is made up of two parts, the point estimate and the margin of error. The point estimate of the difference between two population means is simply the difference between two sample means ($ \bar{x}_1 - \bar{x}_2 $). The standard error of $ \bar{x}_1 - \bar{x}_2 $, which is used in computing the margin of error, is given by the formula below.

The formula for the margin of error depends on whether the population standard deviations ($\sigma_1$ and $\sigma_2$) are known or unknown. If the population standard deviations are known, then they are used in the formula. If they are unknown, then the sample standard deviations ($s_1$ and $s_2$)are used in their place. To change from $\sigma$ known to $\sigma$ unknown, click on $\boxed{σ}$ and select $\boxed{s}$ in the Two Population Calculator.

While the formulas for the margin of error in the two population case are similar to those in the one population case, the formula for the degrees of freedom is quite a bit more complicated. Although this formula does seem intimidating at first sight, there is a shortcut to get the answer faster. Notice that the terms $\frac{s_1^2}{n_1}$ and $\frac{s_2^2}{n_2}$ each repeat twice. The terms are actually computed previously when finding the margin of error so they don't need to be calculated again.

If the two population variances are assumed to be equal, an alternative formula for computing the degrees of freedom is used. It's simply df = n1 + n2 - 2. This is a simple extension of the formula for the one population case. In the one population case the degrees of freedom is given by df = n - 1. If we add up the degrees of freedom for the two samples we would get df = (n1 - 1) + (n2 - 1) = n1 + n2 - 2. This formula gives a pretty good approximation of the more complicated formula above.

Just like in hypothesis tests about a single population mean, there are lower-tail, upper-tail and two tailed tests. However, the null and alternative are slightly different. First of all, instead of having mu on the left side of the equality, we have $\mu_1 - \mu_2$. On the right side of the equality, we don't have $\mu_0$, the hypothesized value of the population mean. Instead we have $D_0$, the hypothesized difference between the population means. To switch from a lower tail test to an upper tail or two-tailed test, click on $\boxed{\geq}$ and select $\boxed{\leq}$ or $\boxed{=}$, respectively.

Again, hypothesis testing for a single population mean is very similar to hypothesis testing for two population means. For a single population mean, the test statistics is the difference between mu and mu0 dividied by the standard error. For two population means, the test statistic is the difference between $\bar{x}_1 - \bar{x}_2$ and $D_0$ divided by the standard error. The procedure after computing the test statistic is identical to the one population case. That is, you proceed with the p-value approach or critical value approach in the same exact way.

The calculator above computes confidence intervals and hypothesis tests for the difference between two population means. The simpler version of this is confidence intervals and hypothesis tests for a single population mean. For confidence intervals about a single population mean, visit the Confidence Interval Calculator . For hypothesis tests about a single population mean, visit the Hypothesis Testing Calculator .

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Hypothesis Test for Population Standard Deviation for normal population

Hypothesis Testing Calculator

Navigating hypothesis testing: unveiling the potential of the hypothesis testing calculator.

Embarking on the journey of statistical exploration, hypothesis testing stands out as an indispensable method for informed decision-making and drawing meaningful conclusions from data. Whether you find yourself in the academic realm, engaged in research endeavors, or navigating the professional landscape, having a trustworthy Hypothesis Testing Calculator in your statistical toolkit can prove to be a game-changer. Let’s delve into the intricacies of hypothesis testing and uncover how this calculator can be your ally in statistical analyses.

Demystifying Hypothesis Testing:

Null Hypothesis (H0): Positioned as the default assumption, the null hypothesis asserts the absence of any significant difference or effect and is commonly represented as H0.

Alternative Hypothesis (Ha): In direct contradiction to the null hypothesis, the alternative hypothesis posits the existence of a noteworthy difference or effect, denoted as Ha.

Significance Level (α): Acting as the predetermined threshold, typically set at 0.05 or 5%, the significance level plays a pivotal role in determining statistical significance. Should the calculated p-value fall below α, the null hypothesis is rejected.

p-value: Representing the likelihood of observing the results, or more extreme outcomes, under the assumption of the null hypothesis being true, a smaller p-value suggests the unlikelihood of the results occurring by chance.

Features that Define the Hypothesis Testing Calculator:

Input Parameters: The calculator demands input of sample data, selection of the test type (e.g., t-test, chi-square test), specification of null and alternative hypotheses, and determination of the significance level.

Calculations: Once armed with the requisite data and parameters, the calculator diligently executes statistical tests and computations. The output encompasses crucial details like the test statistic, degrees of freedom, and the all-important p-value.

Interpretation: Armed with the results, the calculator aids in the decision-making process, guiding whether to reject or accept the null hypothesis. An interpretation of the findings is provided, playing a pivotal role in drawing insightful conclusions.

Visual Representation: Some calculators go the extra mile by offering visual aids such as graphs or charts, facilitating a deeper understanding of data distribution and test outcomes.

Unveiling the Significance of the Hypothesis Testing Calculator:

In Scientific Research: Researchers spanning diverse fields leverage hypothesis testing to validate their hypotheses, thereby extracting meaningful insights from data.

In Quality Control: Industries rely on hypothesis testing as a quality assurance mechanism, ensuring the consistency and excellence of products and processes.

In Medical Studies: Within the realm of medical research, hypothesis testing serves as a critical tool for evaluating the effectiveness of treatments or interventions.

In Academics: Both students and educators find value in hypothesis testing as an educational tool, enabling the comprehension of statistical concepts and the conduct of experiments.

In Data-Driven Decision-Making: Businesses, keen on making decisions grounded in data, turn to hypothesis testing to navigate choices such as launching a new product based on comprehensive market research.

Concluding Insights:

The Hypothesis Testing Calculator emerges as a formidable ally, simplifying intricate statistical analyses and fostering data-driven decision-making. Whether you are in the midst of experimental undertakings, scrutinizing survey data, or overseeing quality control protocols, a solid understanding of hypothesis testing coupled with the use of this calculator empowers you to make well-informed choices. In doing so, you not only contribute to evidence-based research but also play a pivotal role in shaping decision-making processes across various domains.

T-test for two Means – Unknown Population Standard Deviations

Instructions : Use this T-Test Calculator for two Independent Means calculator to conduct a t-test for two population means (\(\mu_1\) and \(\mu_2\)), with unknown population standard deviations. This test apply when you have two-independent samples, and the population standard deviations \(\sigma_1\) and \(\sigma_2\) and not known. Please select the null and alternative hypotheses, type the significance level, the sample means, the sample standard deviations, the sample sizes, and the results of the t-test for two independent samples will be displayed for you:

The T-test for Two Independent Samples

More about the t-test for two means so you can better interpret the output presented above: A t-test for two means with unknown population variances and two independent samples is a hypothesis test that attempts to make a claim about the population means (\(\mu_1\) and \(\mu_2\)).

More specifically, a t-test uses sample information to assess how plausible it is for the population means \(\mu_1\) and \(\mu_2\) to be equal. The test has two non-overlapping hypotheses, the null and the alternative hypothesis.

The null hypothesis is a statement about the population means, specifically the assumption of no effect, and the alternative hypothesis is the complementary hypothesis to the null hypothesis.

Properties of the two sample t-test

The main properties of a two sample t-test for two population means are:

• Depending on our knowledge about the "no effect" situation, the t-test can be two-tailed, left-tailed or right-tailed
• The main principle of hypothesis testing is that the null hypothesis is rejected if the test statistic obtained is sufficiently unlikely under the assumption that the null hypothesis is true
• The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true
• In a hypothesis tests there are two types of errors. Type I error occurs when we reject a true null hypothesis, and the Type II error occurs when we fail to reject a false null hypothesis

How do you compute the t-statistic for the t test for two independent samples?

The formula for a t-statistic for two population means (with two independent samples), with unknown population variances shows us how to calculate t-test with mean and standard deviation and it depends on whether the population variances are assumed to be equal or not. If the population variances are assumed to be unequal, then the formula is:

On the other hand, if the population variances are assumed to be equal, then the formula is:

Normally, the way of knowing whether the population variances must be assumed to be equal or unequal is by using an F-test for equality of variances.

With the above t-statistic, we can compute the corresponding p-value, which allows us to assess whether or not there is a statistically significant difference between two means.

Why is it called t-test for independent samples?

This is because the samples are not related with each other, in a way that the outcomes from one sample are unrelated from the other sample. If the samples are related (for example, you are comparing the answers of husbands and wives, or identical twins), you should use a t-test for paired samples instead .

What if the population standard deviations are known?

The main purpose of this calculator is for comparing two population mean when sigma is unknown for both populations. In case that the population standard deviations are known, then you should use instead this z-test for two means .

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Hypothesis Test for a Mean

Use the calculator below to analyze the results of a hypothesis test for a mean. Enter your null hypothesis's mean, sample mean, sample standard deviation, sample size, test type, and significance level to find your results.

You will find a description of how to conduct a hypothesis test of a mean below the calculator.

Define the t-test

Sample Mean Under the Null Distribution

Conducting single mean hypothesis tests.

A hypothesis test of a sample mean can help you make inferences about the population from which you drew it. It is a tool to determine what is probably true about an event or phenomena.

Testing a Mean

For the results of a hypothesis test of a mean to be valid, you should follow these steps:

Check Your Conditions

State your hypothesis, determine your analysis plan, analyze your sample, interpret your results.

To use the testing procedure described below, you should check the following conditions:

• Simple Random Sampling - You should collect your sample with simple random sampling. This type of sampling requires that every occurrence of a value in a population has an equal chance of being selected when taking a sample.
• Normal Sampling Distribution -The sampling distribution should follow the Normal or a nearly Normal distribution. A sampling distribution will be nearly Normal when the samples are collected independently and when the population distribution is nearly Normal. Generally, the larger the sample size, the more normally distributed the sampling distribution. Additionally, outlier data points can make a distribution less Normal, so if your data contains many outliers, exercise caution when verifying this condition.

You must state a null hypothesis and an alternative hypothesis to conduct a hypothesis test for a mean.

The null hypothesis, is a skeptical claim that you would like to test. It is defined by the null hypothesis's mean, which is often labeled μ 0 .

The alternative hypothesis represents an alternative claim to the null hypothesis.

Your null hypothesis and alternative hypothesis should be stated in one of three mutually exclusive ways listed in the table below.

Before conducting a hypothesis test, you must determine a reasonable significance level, α, or the probability of rejecting the null hypothesis assuming it is true. The lower your significance level, the more confident you can be of the conclusion of your hypothesis test. Common significance levels are 10%, 5%, and 1%.

To evaluate your hypothesis test at the significance level that you set, consider if you are conducting a one or two tail test:

• Two-tail tests divide the rejection region, or critical region, evenly above and below the null distribution, i.e. to the tails of the null sampling distribution. For example, in a two-tail test with a 5% significance level, your rejection region would be the upper and lower 2.5% of the null distribution. An alternative hypothesis of μ ≠ μ 0 requires a two-tail test.
• One-tail tests place the rejection region entirely on one side of the distribution i.e. to the right or left tail of the null sampling distribution. For example, in a one-tail test evaluating if the sampling distribution is above the null sampling distribution with a 5% significance level, your rejection region would be the upper 5% of the null distribution. μ > μ 0 and μ 0 alternative hypotheses require one-tail tests.

The graphical results section of the calculator above shades rejection regions blue.

After checking your conditions, stating your hypothesis, determining your significance level, and collecting your sample, you are ready to analyze your hypothesis.

Sample means follow the Normal Distribution with the following parameters:

• The Population Mean, μ - The population mean is assumed to be the null hypothesis's mean in a single mean hypothesis test.
• The Standard Error, SE - For samples that are much smaller than the population, the standard error can be computed as follows: SE = s / sqrt(n), with s being the sample standard deviation and n being the sample size. It defines how sample means are expected to vary around the null hypothesis's mean given the sample size and under the assumption that the null hypothesis is true.
• Degrees of Freedom, DF - For hypothesis tests for a single mean, the degrees of freedom equals n – 1, with n being the sample size.

In a hypothesis test for a mean, we calculate the probability that we would observe the sample mean, x̄, assuming the null hypothesis is true, also known as the p-value . If the p-value is less than the significance level, then we can reject the null hypothesis.

You can determine a precise p-value using the calculator above, but we can find an estimate of the p-value manually by calculating the t-score, or t-statistic, as follows: t = (x̄ - μ 0 ) / SE

The t-score is a test statistic that tells us how far our observation is from the null hypothesis's mean under the null distribution. Using any t-score table, we can look up the probability of observing the results under the null distribution. You will need to look up the t-score for the type of test you are conducting, i.e. one or two tail. A hypothesis test for a mean is sometimes known as a t-test because of the use of a t-score in analyzing results.

If you find the probability is below the significance level, we reject the null hypothesis.

The conclusion of a hypothesis test for a mean is always either:

• Reject the null hypothesis
• Do not reject the null hypothesis

If you reject the null hypothesis, you cannot say that your sample mean is the true population mean. If you do not reject the null hypothesis, you cannot say that the null hypothesis is true.

A hypothesis test is simply a way to look at a sample and conclude if it provides sufficient evidence to reject the null hypothesis.

Example: Hypothesis Test for a Mean

Let’s say that you manage a clothing store with a historical average transaction amount of $53.24.

You believe the best way to improve your business is to increase your average sale amount. So, you have been working hard on training your sales staff on how sell more items to customers to increase sales.

To test if your sales training has increased your average sale amount, you decide to run a hypothesis test for a mean with a sample of 50 transactions to see if your average sale amount has increased.

• Check the conditions - You collect your sample using simple random sampling , and you know that historically your transactions are normally distributed about the average transaction. So, your conditions for running a hypothesis test for a mean are satisfied.
• State Your Hypothesis - Your null hypothesis is that your average transaction amount is the same or less than the historical average, formally stated μ ≤ $53.24. Your alternative hypothesis is that your average is greater than the historical average, formally stated μ > $53.24.
• Determine Your Analysis Plan - You believe that a 5% significance level is reasonable. As your test is one-tail test, you will evaluate if your sample mean would occur at the upper 5% of the null distribution.
• Analyze Your Sample - After collecting your sample (which you do after steps 1-3), you find that the sample mean, x̄, transaction amount is $61.02 with a standard deviation of $39.12. Using the calculator above, you find that a sample mean of $61.02 would results in a t-score of 1.41 under the null distribution, which translates to a p-value of 8.30%.
• Interpret Your Results - Since your p-value of 8.30% is greater than the significance level of 5%, you do not have sufficient evidence to reject the null hypothesis.

In this example, you found that you cannot reject the claim that your current transaction amount is less than or equal to your historical transaction amount of $53.24. Your results do not guarantee that your transaction amount is $53.24 or below, but they do indicate that your sales training has likely not had the effect that you wanted on your store's average sale amount.

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How to calculate the standard deviation and t-test statistic of matched pairs

So, I have two variables (matched pairs) that I believe are means, with a collection of 10 samples a piece (although if I'm reading it right, they're the same 10 individuals undergoing different trials). It looks a bit like this:

Situation 1: 125, 156, 140, 175, 153, 148, 180, 135, 168, 157 Situation 2: 120, 145, 142, 150, 160, 148, 160, 142, 162, 150

Now the problem is asking for a t test statistic and a p-value, the test statistic is $\frac{\bar x_d}{s_d/\sqrt{n_d}}$ , where:

$\bar x_d$ should be the difference between all of the means summed, so Avg(situation1) - Avg(situation2), which comes out to 5.8.

$s_d$ is the standard deviation (that's all my textbook is giving me).

And $n_d$ is just 10.

Any help would be appreciated, my past two attempts at this problem have been complete bummers - the standard deviation I started trying was 17.512 because that's just what my technology spits out when I put all the values in.

• hypothesis-testing

• 1 $\begingroup$ Please add the self-study tag & read its wiki . $\endgroup$ –  gung - Reinstate Monica Commented Dec 2, 2020 at 19:16
• 1 $\begingroup$ Your task is to transform this apparent two-sample case to a one-sample case, by substracting each value from situation 1 to situation 2 (or the reverse, it doesn't really matter for numerical computation). Let's call this new sample "d". Then, think about how to compute $\bar x_d$ and $s_d$ from "d". $\endgroup$ –  chl Commented Dec 2, 2020 at 19:16
• $\begingroup$ Oh, ok! So, instead of two different stdevs, it's the stdev of the set of values that I get from subtracting the values in situation 2 from situation 1, making 𝑠 10.654 and x 5.8. Super duper helpful, thank you! $\endgroup$ –  Candy H Commented Dec 2, 2020 at 19:31

@chi has given a verbal description. Here is how it looks in the R t.test procedure.

Notice that the t statistic for the t tests below is as follows:

First, doing a one-sample test on the differences d to test the null hypothesis that the population mean of the paired differences is $0.$ The sample mean of the differences is greater $\bar D = 5.8 > 0,$ but compared with the standard deviation $S_d = 10.65$ we cannot say that $\bar D$ is significantly larger than $0.$ The P-value exceeds $0.05 = 5\%,$ so the null hypothesis is not rejected. [The population difference might be anywhere between $-1.822$ and $13.422.]$

Second, if you use the parameter pair=T for t.test , then the program find the differences and does the same thing as above; the t-statistic, degrees of freedom, and P-value are all the same as before.

Notes: (1) In both cases the null hypothesis is that the population of pairs has mean $0$ and the alternative hypothesis is two-sided. (Additional parameters in t.test are required if you want to change either.) (2) Because each pair is for a different individual it is reasonable to assume that there is a positive correlation between x1 and x2 . [Because x1 and x2 are not independent samples, it would be incorrect to do a two-sample t.test.]

Also, t tests are exact only if the data (differences here) are from a normal population. With only $n=10$ differences it is difficult to assess normality of the differences. However, there is no evidence of strong skewness in the stripchart, and there are no extreme outliers. Moreover, a Shapiro-Wilk test shows that differences are consistent with a normal sample.

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Hypotheses testing without a standard deviation

So, there seems to be a gap in my knowledge somewhere... here's the question

A species of a common butterfly with a large population occurs in two different types, light (type $L$) and dark (type $D$). A naturalist observes 10 of the butterflies in a particular wood, $2$ of which are type $L$ and $8$ are type $D$. The naturalist wishes to decide whether this sample provides significant evidence that the proportion of type L butterflies differs from the proportion of type $D$ butterflies in this wood.

Is there evidence to reject the null hypothesis at the $5\%$ level?

So I believe this requires the T distribution due to the small sample size. So using $\frac{X-\mu}{\frac{\sigma}{\sqrt{10}}}$. I take $X$ as $=5$ as it's half the size as the proportion of both $L$ and $D$ should be $=$. Then $\mu=2$ as that was the actual value and $n$ as $10$. But what would $\sigma$ be? Or am I approaching this wrong? is there a way to find the standard deviation from this?

• probability
• probability-distributions
• hypothesis-testing

When testing for proportions, we do not use a $t$-distribution because $p$ is a Bernoulli random variable with only one parameter to estimate since the variance $p(1-p)$ is a function of the mean $p$.

This is a one sample test of proportions where

$$H_0 : p=0.5$$

$$H_a : p\neq 0.5$$

$$Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$$

In our case, we can use $\hat{p}=0.8$ or $\hat{p}=0.2$ and come to the same conclusion.

$$Z=\frac{0.8-0.5}{\sqrt{\frac{0.5(1-0.5)}{10}}}=1.897$$

By a $z$-table, you should find

$$P(|Z|\gt 1.897)\approx 2\cdot0.0287=0.0574$$

R statistical software give a p-value of $0.0578$

Thus, at $\alpha=0.05$ we fail to reject the null hypothesis that $p=0.5$ so we do not have significant evidence that the proportion of type L butterflies differs from the proportion of type D butterflies in this population.

Non-Parametric Test:

You can also run a binomial test to get the exact probability. Let $X$ denote the number of type D butterflies we observe. Then

$$P(X\leq 2 \cup X \geq 8 \mid p=0.5)\approx 0.1094$$

Here we fail to reject the null hypothesis as well.

• $\begingroup$ So you're saying we use the normal distribution table? Cos I thought we needed a large pool of data for that, say $>30$? and would we then do the same if we had looked at 1000 butterflies as with 10? $\endgroup$ –  Dan D'silva Commented Apr 16, 2018 at 16:41
• $\begingroup$ Yes, when testing for proportions, we always use a normal distribution. Never a $t$-distribution. No matter the sample size. $\endgroup$ –  Remy Commented Apr 16, 2018 at 16:54
• $\begingroup$ One last thing. Why at the end do we times it by 2? And is this something I should be doing normally? I'm assuming its something to do with both tails of the distribution. could you explain this? $\endgroup$ –  Dan D'silva Commented Apr 17, 2018 at 14:54
• $\begingroup$ Because it's a two-tailed test. When the alternative hypothesis is $\lt$ or $\gt$ then you don't multiply by two. $\endgroup$ –  Remy Commented Apr 17, 2018 at 17:48

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Understanding Advanced Tableau Calculations Like Standard Deviation, Moving Average, and More

Maddie Dierkes

• Decision Science
• Engineering
• Tableau Desktop
• Technical Features

Tableau is a useful tool for an analyst to have in their repository, with its drag and drop interface allowing practitioners to easily dive in to explore their data or spend time creating sustainable, decision-centric dashboards. Beyond its user experience and user interface capabilities, Tableau also provides native calculations that have the versatility to integrate more complex calculations into your analyses. In this post, I am going to take you through a few of those, why they are important, and how to execute them in Tableau. We will cover weighted averages, Tableau moving averages, Tableau standard deviations, and Tableau percentiles using the Sample – Superstore dataset.

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Weighted averages

A weighted average is simple to build in Tableau. All you are doing is taking the sum of the value you want to average (i.e. Profit), multiplying by the weighting value (i.e. Amount) and then dividing it by the total of the weighting value (Amount, in this case).

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Weighting becomes important as you look at your data at a higher level of grain than the smallest level. If every row in a dataset represented one individual sale, and sales was the weighting amount, then your weighted average would equal the straight average. Instead of working with a dataset where each row is an individual sale, what if your data was already aggregated to the state level with the average profit per order and amount of orders. 

Why It’s Important to Understand the Granularity of Data

Imagine your task is to show the overall profit per order. Your mind may want to simply average the column to the left, essentially taking an average of the average. You could do that if your data was at an order number level like Sample – Superstore. But if your data came in aggregated at the state level, a weighted average provides a more accurate representation of the data.  

To compute the weighted average, multiply the Amount by the Profit Per Order for each state and sum it up. Take that value and divide it by the sum of the Amount :

SUM([Amount]*[Avg. Profit Per Order])/SUM([Amount])

You get nearly a $50 difference when comparing this result to a straight average.

Reporting a straight average out to stakeholders would misrepresent the profit and could lead to misinformed decision making. This technique reduces the impact of the states with less sales and allows states with higher sales to have a higher impact in the overall average.

Tableau standard deviation

Standard deviation is a statistical measure of how dispersed the data is and is used in hypothesis testing and outlier detection . There are two equations for standard deviation, Population Standard Deviation and Sample Standard Deviation.

Population should be used when you are conducting a test on an entire population while the sample equation should be used for a subset. The variables for the equations are as follows:

• x i is an individual value in the data
• u is the population mean
• N is the population size
• x is the sample mean
• n is the sample size

For each calculation, it takes a sum of each individual value in the data, subtracts the mean from it, and then squares it. Then it is divided by either the population size or the sample size minus 1 and the square root is taken of that value.

Standard deviations are already built in Tableau functions. STDEV() for sample and STDEVP() for population are built in calculations that allow you to easily calculate the standard deviation in Tableau for whatever dimensions you have on your view. If I needed to calculate the standard deviation of Profit in Tableau for the entire Superstore data set, I would use STDEVP() which results in $232.45.

STDEVP([Profit])

STDEV([Profit])

Tableau moving average

A moving average is another reporting technique that smooths out data and reduces the impact of short-term volatility in the data. There are two types of moving averages: simple and exponential . Simple moving averages are calculated by looking back a predetermined distance in the data and taking the average up through the current data point. An exponential moving average gives more weight to more recent values.

A simple moving average can be calculated in Tableau utilizing the WINDOW_AVG table calculation. In this example, we are going to say that we want to create a moving average of the monthly profit looking back 3 months and including the current month in the average. I am going to create a parameter to represent the number of months we want to average. 

WINDOW_AVG(SUM([Profit]),-([Period]-1),0)

This calculation is averaging the sum of profit for the previous two months and the current month. For UX purposes, I have set the Period parameter to three, so I will need to subtract one from the period so you average the current month with the previous 2 months, creating a 3-month moving average.

The Beginner’s Guide to Tableau Table Calculations

Create a line chart by dragging the Order Date field to the Columns shelf, setting the date format as continuous Month. Drag Profit and Moving Average to the Rows shelf, create a dual axis, and synchronize the axes. Comparing the Tableau moving average of the profit sum to just the profit sum month by month, you get a less volatile look at the trend:

Next, let us take a look at another variation of a moving average, the exponential moving average . Instead of a straight average of the previous months, it gives more weight to more recent months. 

The equation for an exponential moving average is:

Current Value is the actual value for the specified time period, Previous EMA is the value for the last data point. The 𝞪 is a calculated value based on the number of periods you want to look backward. 

We can repurpose the Period parameter we created for a Tableau simple moving average, so now we need to create the 𝞪 calculation in Tableau. In mathematical notation it looks like:

Where n is the number of time periods. In our example we have it set to three. In Tableau it looks like:

2/([Period]+1)

With our Alpha value, we can now create our EMA function in Tableau:

([Alpha]*SUM([Profit]))+(PREVIOUS_VALUE([Moving Average])*(1-[Alpha]))

This equation is taking the Alpha and multiplying by the current period’s value. Then adding that to the previous EMA value multiplied by one minus the alpha value. The PREVIOUS_VALUE function is a helpful table calculation that allows us to avoid a circular reference. It pulls the previous value of the function you are currently creating. If there is no previous value (which would be true for the first month in the view), it will pull the current sum of the profit. 

Create a view like we did before, only this time use the exponential moving average instead of the simple moving average:

The results are very similar so you may be asking which one to use. That is entirely dependent on your use case, but a good example is stock trading. Short term strategies tend to favor the exponential moving average because it focuses more closely on recent history, while long term strategies may want to use a simple moving average for the opposite reason. 

Tableau percentiles

Calculating percentiles gives another layer of context to your data. A percentile is a measure that notes the position of a specific data point relative to the rest of the data. For example, say you got a test score of 70%. The initial reaction of someone who has no insight on the difficulty of the test would assume that is a low score. However, say that you scored in the 95th percentile of students who took the test, meaning you scored higher than 95% of students. Now the initial reaction is that the score was top of the class, and you may assume that the test or class is a very difficult one. 

Percentiles are not too difficult to understand conceptually. To calculate the percentile for a data, point the equation is:

To calculate this, you just need the number of data points with the lesser value. So, say there were 15 students in the class. You scored a 75% on the test and there were 14 students who scored lower than a 75%. The percentile would be calculated at (14/15)*100, so your score would be in the 93.3 percentile or rounding down to the 93rd percentile.

Tableau makes it easy for the user with built in percentile functions. There are four options for Tableau Percentiles

• RANK_PERCENTILE()
• PERCENTILE()
• WINDOW_PERCENTILE()
• MODEL_PERCENTILE()

In this post, we will only be covering RANK_PERCENTILE() and PERCENTILE().

First, let’s look at the RANK_ PERCENTILE() function. RANK_PERCENTILE is a table calculation that will return the percentile for each point in the data. I want to calculate the percentile for the total sales by customer. The calculation is as follows:

RANK_PERCENTILE(SUM([Sales]))

There is also the option to drag the sum of sales into the view and do a Quick Table Calculation, choosing percentile as the table calculation. I am going to create a simple dot plot of all the customers and add Rank_Percentile to the tooltip. Since this is a table calculation, I am going to make sure that the table calculation is set to calculate across Customer Name

Doing an analysis, this dot plot is extremely crowded. If I click on a customer in the middle of the chart, instinct would lead me to believe that the customer falls somewhere close to the 50th percentile in terms of sales. However, Ben Ferrer actually sits in the 96th percentile, having total sales higher than 96% of the other customers even though he sits in the middle of the chart.

The Percentile() function does the opposite, where instead of finding the percentile for each value it finds the value for each percentile. The equation is as follows in Tableau:

PERCENTILE([Sales]),.75)

This will return the value where 75% of sales fall below at the most granular level in the data. For my analysis, I wanted to add in two level of detail calculations: one to fix the sum of sales to Customer Name and the other to fix the percentile value to be calculated across the entire dataset. 

{PERCENTILE({Fixed [Customer Name]: SUM([Sales])},.75)}

I am going to then create a Boolean calculation to determine if a value is greater than the 75th percentile value. Then I will add this as a color to my chart to color all dots whose sales value is less than the 75th percentile. 

As you can see, the 75th percentile falls visually about ⅓ up the chart and that there is a greater dispersion amongst the top 25% of customers.

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Let’s stay in touch:

To wrap up, I am going to put together a quick dashboard analyzing sales data in the Sample – Superstore dataset using some of the calculations we touched on.

These calculations allow you to begin answering some of the questions stakeholders might have, taking a dashboard from just a visual to a decision-making tool. Now you should have a good jumping off point to begin incorporating these into your analyses! 

Stay after it, Maddie

Peer Review

Director, Analytics Engineering

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The psychology department is gradually changing its curriculum by increasing the number of online course offerings. To evaluate the effectiveness of this change, a random sample of n=36 students who registered for Introductory Psychology is placed in the online version of the course. At the end of the semester, ali students take the same final exam. The average score for the sample is M=76. For the general population of students taking the traditional lecture class, the final exam scores form a normal distribution with a mean of μ=71. If the final exam scores for the population have a standard deviation of a=12, does the sample provide enough evidence to conclude that the new online course is significantly different from the traditional class? Use a two-talled test with a=.05. z-critical =± z= These results indicate: Rejection of the null hypothesis; the course has a significant effect Rejection of the null hypothesis; the course does not have a significant effect Failure to reject the null hypothesis; the course does not have a significant effect Fallure to reject the nuil hypothesis; the course has a significant effect If the final exam scores for the population have a standard deviation of σ=18, does the sample provide enough evidence to conclude that the new online course is significantly different from the traditional class? Again, use a two-tailed test with a =.05. (Round z score to two decimal places.) z-critical =± z These results indicate: Failure to reject the null hypothesis; the course has a significant effect Rejection of the null hypothesis; the course has a significant effect Failure to reject the null hypothesis; the course does not have a significant effect Rejection of the null hypothesis; the course does not have a significant effect Comparing your answers for the two cases, explain how the magnitude of the standard deviation influences the outcome of a hypothesis test. A larger standard deviation produces a larger standard error, which reduces the likelihood of rejecting the null hypothesis. A smaller standard deviation produces a smaller standard error, which reduces the likelihood of ickecting the null hypothesis. A larger standard deviation produces a smaller standard error, which reduces the likelihood of rejecting the null hypothesis. A smaller standard devistion produces a larger standard error, which reduces the likelihood of rejecting the null hypothesis,

The magnitude of the standard deviation affects the likelihood of rejecting the null hypothesis in a hypothesis test. A larger standard deviation decreases the likelihood of rejecting the null hypothesis, while a smaller standard deviation increases the likelihood.

The magnitude of the standard deviation affects the likelihood of rejecting the null hypothesis.

The standard deviation plays a crucial role in hypothesis testing. When conducting a hypothesis test, the standard deviation is used to calculate the standard error, which measures the variability of sample means. A larger standard deviation leads to a larger standard error, indicating that the sample means are more spread out from the population mean. Consequently, it becomes less likely to reject the null hypothesis because the observed differences between the sample and population means may be attributed to random variation rather than a true effect.

On the other hand, a smaller standard deviation results in a smaller standard error, indicating that the sample means are more tightly clustered around the population mean. In this case, even small differences between the sample and population means are more likely to be statistically significant , increasing the chances of rejecting the null hypothesis.

In the first scenario where the standard deviation of the population is 12, the sample's average score of 76 is significantly different from the population mean of 71. This leads to the rejection of the null hypothesis, suggesting that the new online course has a significant effect on the final exam scores.

In the second scenario where the standard deviation of the population is 18, the sample's average score of 76 is not significantly different from the population mean of 71. Therefore, the null hypothesis is not rejected, indicating that there is not enough evidence to conclude that the new online course is significantly different from the traditional class.

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(organizational behavior) in reflecting on the following problem situation, consider (1) the strengths and weaknesses of the person's type, (2) the demands of the work environment, and (3) the managerial approaches to problem solving that you think should result from an understanding of personality type. Situation: An ISFJ employee is having problems making face-to-face, in-person sales calls with completely new clients who have not previodsly contacted the company (cold calls) and in negotiating contracts with customers in person. How could you work with this person to make him or her successful in this role?

Supporting an ISFJ employee in making successful cold calls and negotiations involves recognizing their strengths and weaknesses, providing a supportive work environment, and using positive and trust-building managerial approaches to problem-solving.

As an ISFJ employee is having difficulties making cold calls to new clients and negotiating contracts in person, here is how you could work with him/her to make him/her successful in this role: 1. ISFJ Strengths and WeaknessesISFJ is a type of personality that is loyal, empathetic, and detail-oriented.

However, they can also be hesitant, risk-averse, and reluctant to engage with new people in a new setting. To make them successful in their role, it is important to recognize these strengths and weaknesses.

2. Work Environment Demands - Sales and negotiating contracts are both activities that necessitate persuasive abilities, assertiveness, and self-confidence. It is crucial to provide an environment that is supportive and encourages the development of these abilities.

This may entail providing a mentor who is more confident and assertive than the ISFJ employee, as well as training that focuses on soft skills and personal growth. 3. Managerial Approaches to Problem SolvingWhen dealing with an ISFJ employee, it is important to maintain a positive attitude and avoid confrontational techniques.

Instead, build trust with the employee by establishing an open and supportive working relationship. Encourage the employee to take risks by providing opportunities for growth and development. For instance, you can try role-playing to help them feel more comfortable in the process.

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a. Explanation of Lee's letter I. b. Explanation of Federalist 84.

Lee's Letter I expressed concerns about the proposed Constitution, emphasizing the need for a Bill of Rights, while Federalist 84, written by Hamilton, argued against the inclusion of a Bill of Rights, highlighting the existing protections within the Constitution. The debate led to the eventual adoption of the Bill of Rights.

a. Lee's Letter I refers to the first of a series of letters written by Richard Henry Lee in 1787. Lee was a prominent Antifederalist who expressed his concerns about the proposed Constitution of the United States, which was being considered for ratification at the time.

In his letter, Lee argued against the consolidation of power and the absence of a Bill of Rights in the Constitution. He believed that the central government would become too powerful and encroach upon the rights of the states and individuals.

Lee emphasized the importance of limiting the scope of the federal government and ensuring that the rights of citizens were protected through explicit guarantees. His letter was influential in sparking the debate over the need for a Bill of Rights, which ultimately led to the adoption of the first ten amendments to the Constitution.

b. Federalist 84 is one of the essays written by Alexander Hamilton as part of The Federalist Papers, a collection of articles published in 1787-1788 to promote the ratification of the United States Constitution. In Federalist 84, Hamilton argued against the inclusion of a Bill of Rights in the Constitution.

He contended that the Constitution, as it stood, already provided sufficient safeguards for individual liberties and that a specific enumeration of rights could be problematic.

Hamilton believed that a Bill of Rights might imply that the government had the power to infringe on rights not explicitly listed, potentially undermining the very liberties it sought to protect.

He also pointed out that the Constitution contained checks and balances, separation of powers, and a system of limited government that would prevent the abuse of individual rights.

Hamilton's arguments were met with opposition from Antifederalists who advocated for a Bill of Rights, but his views ultimately did not prevail, and the Bill of Rights was added to the Constitution in 1791.

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The potential fallacies in this statement include false comparison, misleading generalization, confirmation bias, and neglecting prevention and surveillance.

The statement commits a false comparison fallacy by comparing the situation in developing countries with that in the United States. It assumes that because developing countries face more waterborne disease issues, the United States is completely free from such concerns. However, each country's situation must be assessed independently.

Furthermore, the statement makes a misleading generalization by concluding that waterborne diseases are a non-issue in the United States based solely on the reported deaths from outbreaks. This overlooks potential cases of illness, health impacts, and other consequences that may arise from waterborne diseases, even if they don't result in deaths.

Confirmation bias is evident in the statement as it selectively focuses on the data from the CDC that supports the argument while disregarding other relevant factors or sources of information. This biased perspective can lead to a distorted understanding of the actual risks and challenges associated with waterborne diseases in the United States.

Lastly, the statement neglects the ongoing efforts in the United States to prevent and manage waterborne diseases . Just because there are few reported deaths does not mean that there are no preventive measures in place. The CDC and other public health agencies actively work to ensure water source safety, implement surveillance systems, and educate the public. Ignoring these efforts underestimates the importance of maintaining vigilance and preventive measures in addressing waterborne diseases in the United States.

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The Black Lives Matter (BLM) movement did not use videos, quotes, or staged events to purposefully demonize the police. It is essential to approach this topic with accuracy and fairness .

The BLM movement aims to address systemic racism and advocate for the fair treatment of Black individuals by law enforcement . While the movement has highlighted instances of police violence and misconduct through videos and quotes, the intention is to raise awareness and prompt necessary reforms rather than demonizing the entire police force.

Videos : The circulation of videos capturing incidents of police violence against Black individuals has played a significant role in shedding light on systemic issues. These videos often serve as evidence and documentation of instances that require scrutiny and justice. They have sparked conversations about police accountability and the need for reform.

Quotes : Quotes from activists and supporters of the BLM movement are typically aimed at highlighting concerns about racial profiling, discriminatory practices, and excessive use of force. These quotes serve as powerful statements to draw attention to systemic issues and advocate for change. However, they do not imply a general demonization of all police officers.

Staged Events : Staged events are not a common practice within the BLM movement. The movement primarily focuses on peaceful protests, community organizing, and policy advocacy to address racial injustice and police brutality. It is important to note that individual actions within any movement do not represent the entire movement as a whole.

It is crucial to approach discussions about social movements with a fair and nuanced perspective, avoiding generalizations or mischaracterizations . The BLM movement seeks to address systemic issues rather than demonize the police as a whole. It advocates for accountability, justice, and equality for all individuals within the criminal justice system.

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Which of the following is true for AI? a. AI-driven processes are less scalable than traditional processes; b. AI-driven processes enable greater scope, as they easily connect with a myriad of other digitized businesses; c. AI-driven processes deliver less powerful opportunities in learning, such as producing accurate, complex, and sophisticated predictions; d. AI cannot change how industries function and the way the economy operates.

The following is true for AI is B. AI-driven processes enable greater scope, as they easily connect with a myriad of other digitized businesses.

Artificial intelligence has rapidly transformed the modern business environment by providing businesses with new and innovative tools to boost their operations. Artificial intelligence (AI) is a computer system that learns and executes tasks that would usually require human intelligence. With the rapid expansion of data analytics and computing power, AI has been a growing trend, with businesses starting to implement it into their operations.

The implementation of AI-powered solutions has the potential to revolutionize how businesses operate, from streamlining tasks and decision-making to providing advanced and more accurate predictions for future events. The AI-driven process also enables greater scope, as they easily connect with a myriad of other digitized businesses. As AI systems are often connected to cloud-based systems, they can quickly and easily integrate with other AI-powered solutions, increasing the speed and efficiency of their operations. Thus, the correct answer is B. AI-driven processes enable greater scope, as they easily connect with a myriad of other digitized businesses.

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The specific impact of COVID-19 on unemployment in Canada can only be accurately assessed through the analysis of up-to-date data and statistics.

COVID-19 has had a significant impact on the global economy , and Canada has not been immune to its effects. The pandemic has resulted in widespread business closures, disruptions in supply chains, and reduced consumer spending, leading to job losses across various sectors.

The impact of COVID-19 on unemployment in Canada is likely to vary across provinces due to differences in economic structure and industry composition. Provinces heavily reliant on sectors such as tourism, hospitality , and retail may experience higher unemployment rates compared to provinces with a more diverse economic base. Provinces with a strong healthcare or technology sector may be more resilient.

Furthermore, gender disparities in employment have been observed in many countries, including Canada. The pandemic's effects on unemployment can disproportionately affect certain industries that employ more women, such as hospitality and retail. Women may also face additional challenges due to caregiving responsibilities and the burden of unpaid work at home.

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) With grades, universities and subjects seem to evaluate the quality of their own teaching. Universities and subjects compete for good students, and society rewards them for their teaching results. Students also hope for good grades. Universities and subjects have a lower price for good grades. The result is that the average grades in universities and subjects have increased over time, and their informational value has weakened. phenomenon known as grade inflation. Select one or more: grade inflation is an example from the central limit value theorem Goodman's Law From the Hawthorn effect none of the previous ones. b) Expert knowledge must be assessed in order to be able to: remember, understand, apply, analyze, create, create on a scale. Choose the correct option: scale wanted to be 1 ordinal scale 2 distance scale 3 ratio scale c) Sampson and Winter (2018) study children's exposure to lead, it becomes the next juvenile delinquency in children and argue that environmental policy is also criminal policy. In the data, the sample mean and standard deviation of the blood lead concentration of 212 Chicago children were 6.20 μg/dl and 4.58 μg/dl. Answer assuming that these indicators hold exactly in the population. Can the distribution of a child's blood lead concentration be normally distributed? (Yes or No) (Hint1: Calculate the value of the accumulation function of the N(6.20,4.582) distribution at point 0 with the command pnorm(0,6.20,4.58). Hint2: Can the lead concentration be a negative number?)

The phenomenon described is known as grade inflation . Grade inflation is the increase in average grades over time, resulting in the weakening of their informational value. The correct option for assessing expert knowledge is: 3) ratio scale Regarding the distribution of a child's blood lead concentration No, the distribution cannot be normally distributed.

It occurs when grades are progressively increasing and making it difficult to differentiate between students’ performance . Hence, the average grades in universities and subjects have increased over time, and their informational value has weakened.

b) Ratio scale is a variable measurement scale that satisfies the characteristics of the interval scale with the additional feature of having an absolute zero point. In the ratio scale, ratios of numbers do make sense, and zero on the scale means the complete absence of the variable being measured.

Hence, the correct option is the Ratio scale. c) Yes, the distribution of a child's blood lead concentration can be normally distributed. If the mean and standard deviation of blood lead concentration data is known and the given indicators hold exactly in the population.

Then the normal distribution of blood lead concentration can be found with mean = 6.20 μg/dl and standard deviation = 4.58 μg/dl. Also, blood lead concentration cannot be a negative number as it is a physical quantity that always has a value greater than or equal to zero.

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In how to survive a plague, why weren’t many activists satisfied with azt? group of answer choices it was prohibitively expensive it had toxic side effects it was largely ineffective all the options

Many activists were dissatisfied with AZT due to its toxic side effects and prohibitively expensive cost, as well as their belief that the drug had marginal benefits and was prioritized over other potentially more promising therapies.

Many activists were dissatisfied with AZT for several reasons. Firstly, the drug had toxic side effects that made it unsuitable for some patients. Secondly, AZT was prohibitively expensive, which meant that many people couldn't afford to access the medication . These factors led to frustration among activists who believed that access to effective and affordable treatment should be a basic right for all individuals with HIV/AIDS.

The dissatisfaction with AZT was also fueled by the perception that the drug's benefits were only marginal. Some activists felt that the government was promoting AZT at the expense of other potentially more promising therapies . They called for better research and more efficient testing methods to expedite the approval of alternative drugs. ACT UP, a prominent American political group formed in response to the AIDS epidemic, played a crucial role in advocating for these changes.

In summary, activists expressed dissatisfaction with AZT due to its toxic side effects and high cost. They believed that the drug's benefits were limited and that alternative treatments should be explored more vigorously. Access to affordable and effective medication was a key demand put forth by activists in their efforts to combat the HIV/AIDS epidemic. Therefore, all the options - toxic side effects and prohibitive cost - were valid reasons for their dissatisfaction.

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In studying his campaign plans, Mr. Singleton wishes to estimate the difference between younger voters and older voters views regarding his appeal as a candidate. He asks his campaign manager to take two random independent samples and find the 90% confidence interval for the difference. A random sample of 707 voters age 35 and under and 745 voters over age 35 was taken. 170 voters age 35 and under and 358 voters over age 35 favored Mr. Singleton as a candidate. Find this confidence interval. Find the point estimate that should be used in constructing the confidence interval. Round your answer to three decimal places

The confidence interval for the difference between younger voters and older voters views regarding Mr. Singleton as a candidate is (0.157, 0.239). The point estimate that should be used in constructing the confidence interval is 0.198.

The point estimate is given by:

p1−p2=n1p1+n2p2n1+n2=(707 × 170 + 745 × 358)/(707 + 745) = 0.198

Thus, the point estimate that should be used in constructing the confidence interval is 0.198.

To find the confidence interval , the standard error of the difference between two population proportions should be calculated as follows:

SE=√(p1(1−p1)n1+p2(1−p2)n2n1n2)=√((0.24(1−0.24)707+0.48(1−0.48)745)/707 × 745)=0.022

The 90% confidence interval can be computed using the following formula:

Lower Limit = (p1 − p2) − Zα/2 x SE

Upper Limit = (p1 − p2) + Zα/2 x SE

Lower Limit = (0.24 − 0.48) − 1.645 x 0.022 = 0.157

Upper Limit = (0.24 − 0.48) + 1.645 x 0.022 = 0.239

The confidence interval for the difference between younger voters and older voters views regarding Mr. Singleton as a candidate is (0.157, 0.239).

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An educational consultant has done a survey of students who obtained bachelor's degrees in Anthropology, History and Philosophy regarding their choice for a Master's degree program. The data are given in the following table. Do you think that there is an associative relationship between a student's bachelor's degree field and the choice of a master's program? Do a χ 2 test at 1% level of significance to check if the choice of the master's degree program is independent of the student's field of study in the bachelor's degree program. Refer to the Problem in question 23 and the Table with 6 Columns. What is the value of d, the difference between the observed frequency b and the expected frequency e for cell (2,2) ? Note: d=(b−e) Refer to the Problem in question 23 and the Table with 6 Columns. What is the value of d, the difference between the observed frequency b and the expected frequency e for cell (2,3) ? Note: d=(b−e) Refer to the Problem in question 23 and the Table with 6 Columns. What is the value of d, the difference between the observed frequency b and the expected frequency e for cell (3,1) ? Note: d=(b−e) Refer to the Problem in question 23 and the Table with 6 Columns. What is the value of d, the difference between the observed frequency b and the expected frequency e for cell (3,2) ? Note: d=(b−e) Refer to the Problem in question 23 and the Table with 6 Columns. What is the value of d, the difference between the observed frequency b and the expected frequency e for cell (3,3) ? Note: d=(b−e)

In the given problem, the educational consultant has conducted a survey of students who obtained bachelor's degrees in Anthropology, History and Philosophy regarding their choice for a Master's degree program.

The data is presented in the table given below:Do you think that there is an associative relationship between a student's bachelor's degree field and the choice of a master's program?To check whether there is an associative relationship between a student's bachelor's degree field and the choice of a master's program or not, we have to perform the chi-square (χ²) test of independence.Hypothesis: H0: There is no relationship between the bachelor's degree field and the choice of a master's program.

Ha: There is a relationship between the bachelor's degree field and the choice of a master's program. Level of significance: α = 0.01Degrees of freedom: df = (r - 1) × (c - 1) = (3 - 1) × (3 - 1) = 4where, r = number of rows = 3 and c = number of columns = 3The expected frequency of each cell can be calculated as:ei j​= (Row i total) × (Column j total) / nwhere, n = total number of observationsThe observed frequency b and the expected frequency e for cell (3,3) is as follows:Observed frequency in cell (3,3), b = 80Expected frequency in cell (3,3), e = (180 × 100) / 600 = 30Hence, the value of d, the difference between the observed frequency b and the expected frequency e for cell (3,3) is:d = b - e = 80 - 30 = 50

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Is this a good argument thesis if not how can I rewite it better? Topic: How latina Women face violence in both the US-Mexico border seeking a better life. Argumentative Thesis: The violence Latina women face evey day at the US-Mexico border has been hidden or desensitized for years. Because the majority of the public have been disconnected from the violent truth of the true cruelty women face.

Revised Argumentative Thesis : The pervasive violence inflicted upon Latina women at the US-Mexico border, which has long been overlooked and normalized, perpetuates a cycle of abuse and dehumanization due to a disconnect between the general public and the harsh realities they endure daily.

Your thesis statement highlights an important issue regarding the violence Latina women face at the US-Mexico border. However, it could be further refined to make it more precise and focused. Here's a revised version:

Revised Argumentative Thesis: The pervasive violence inflicted upon Latina women at the US-Mexico border has long been overlooked and normalized, perpetuating a cycle of abuse and dehumanization. This issue persists due to a widespread disconnect between the general public and the harsh realities these women endure daily.

Key improvements in this revised thesis include:

Specificity : By emphasizing the "pervasive violence," you highlight the widespread nature of the issue faced by Latina women.

Clarity: The phrase "overlooked and normalized" brings attention to the fact that this violence has been downplayed or accepted as a norm, reinforcing the need for change.

Impact: Mentioning the perpetuation of a "cycle of abuse and dehumanization" underscores the long-term consequences of this violence.

Audience connection : Referring to the "disconnect between the general public" highlights the broader societal issue, suggesting the need for increased awareness and empathy.

Remember, your thesis serves as a roadmap for your argument, so it's crucial to state your main points clearly and concisely while capturing the essence of your intended argument.

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In terms of privatisation, describe a public-private partnership (PPP). Provide an example of who may be involved in a PPP and what might be the expected benefits from a PPP to the various stakeholder

A public-private partnership (PPP) is a cooperative arrangement between a government or public sector entity and a private company or consortium to jointly undertake a project or provide a public service. In a PPP, both parties share the risks, responsibilities, and rewards associated with the project or service.

An example of a PPP could be the construction and operation of a toll road. The government may partner with a private infrastructure company to design, build, finance, and maintain the road. The private company would invest capital, expertise , and resources in exchange for the right to operate the road and collect toll revenues for a specified period.

The expected benefits of a PPP can vary for different stakeholders:

1. Government/Public Sector : The government can leverage private sector expertise, resources, and funding to develop infrastructure or deliver public services without incurring the full financial burden. It allows the government to share risks, reduce upfront costs, and transfer certain responsibilities to the private partner.

2. Private Company/Consortium : The private partner benefits from revenue generation through user fees or other income streams. They may gain long-term operational and maintenance contracts, providing a steady income stream. Additionally, PPPs offer opportunities for private entities to access new markets, expand their portfolios, and showcase their capabilities.

3. Citizens and Users : PPPs can lead to improved infrastructure and public services, such as transportation networks, healthcare facilities, or water treatment plants. This can enhance the quality of life for citizens by providing better access to essential services, reducing travel time, and improving overall efficiency.

4. Economy and Employment : PPPs can stimulate economic growth and create job opportunities. Infrastructure development projects, for instance, can generate employment during the construction phase and support economic activities in the surrounding areas once operational.

Overall, a well-structured PPP can combine the strengths of both the public and private sectors to deliver efficient and sustainable projects or services. However, it is important to carefully consider the terms of the partnership, ensure transparency, address potential conflicts of interest, and monitor performance to maximize the benefits for all stakeholders.

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write a eulogy for the trade union leader Clement Payne which includes the following aspect of his life; 1. Early life and education 2. Entry into trade unionism 3. overall contribution to the labour movement of jamaica and the possible impact on the Caribbean industrial relations

Ladies and gentlemen, we are gathered here today to honor the memory of a great Jamaican trade union leader, Clement Payne.

Born in Trinidad in 1904, Payne's family later moved to Barbados where he grew up and received his early education. Payne began his involvement in trade unionism in 1931 when he founded the Barbados Workers' Union (BWU) with the aim of improving the conditions of the working class. He was then elected president of the BWU, a position he held for several years. During his time as a trade unionist, Clement Payne made several contributions to the labor movement of Jamaica and the Caribbean as a whole. One of his most significant contributions was his leading role in organizing the labor riots in Jamaica in 1938. This event led to the establishment of trade unions and paved the way for better working conditions for Jamaican workers.

Payne's impact on the Caribbean industrial relations cannot be overstated. He was an inspiration to many of the Caribbean's labor leaders, and his ideas and principles still resonate today. He believed that workers' rights were human rights, and he fought tirelessly to improve the conditions of the working class throughout his life.Clement Payne was a visionary leader, a passionate advocate for workers' rights, and a true hero of the Caribbean labor movement. His legacy will continue to inspire generations of trade unionists and labor leaders to come. We will always remember him as a dedicated and fearless champion of the working class .

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This analyses for this assignment are based on the same data set we used for SPSS assignment #1. Background information: Buss and Perry (1992) developed the Aggression Questionnaire to assess different types of aggression. They created items to measure aggression proneness for four domains of aggression: physical aggression, verbal aggression, anger, and hostility (a combination of resentment and suspicion). Research Question: Your statistics instructor wants to use data from a previous statistics class to determine whether there is a gender difference in self-reports of physical aggression proneness among college students. In other words, is the gender difference in scale scores for physical aggression proneness (PHYSAGG) statistically significant? The SPSS output below is provided to help you answer the questions in Blackboard. Independent Samples Test Levene's Test for Equality of Varianras Hest for Equality of Means Which iest statistie should be used for this research question, and why? A single-sample z4est because the sample was taken from a broaser student population. An independent-samples. F-lsst because each participant was tested trice : An independent-samples t-test, because we are conplaring two independent groups and we don't know any population information. A single-sample t.test because we have one sample of participants who were spit into two groups for comparison. QUESTION 2 The roteach question of interest in this scenario is whether there is a gender difference in physical aggression proneness. is this an exampie of a frue experiment or quasi-experiment? The is an example of a true expermest beciuse there is no manipuloton of the indegendent vorahin and no ranfom assigrment to condions. This is an example of a quasi-experiment because oender was manipidaled by the researcher. This is an example of a true experiment because gender was maniculated by the rebearchet and the panticpants were randon. QUESTION 3 lantiy the indepensent (or quasivindependent) variable in this sitaation. Womea Man Gender Ala'estion Proneness How was the dependent variable operationally defined? Choose the best answer. Physical Aggression Scores on the physical aggression subscale of The Aggression Questionnaire. Aggression Proneness The instructor's observations of physical aggression proneness QUESTION 5 What is the alternative hypothesis for this research situation? H 0 ​ :μ 1 ​ =μ 2 ​ H 1: ​ μ 1 ​  =μ 2 ​ H 1 ​ :μ 1 ​ <μ 2 ​ and H 1 ​ :μ 1 ​ >μ 2 ​ There is no atternative hypothesis. According to the output, how would you report the probability of a Type I error (for the t-test itself, not the test for the homogeneity of variance assumption)? Phrased differently, what was the probability that the mean difference was due to sampling error? (Choose the option that best reflocts how this value should be reported in APA format.) p<.001 p=.310 p=.000 p=.094 ​ QUESTION 8 Make a decision regarding the null hypothesis. What is the rationale for your decision? Select all of the following that are true regarding your decision about the mull hypothesis, how you made this decision, and what the implications of the decision are. Women had higher scores on physical aggression proneness than men did. The correct decaion in this research situaton is to tall to reject the null hypothesis. The correct becision in this research situation is to reject the null hypotheeis. The resulis are statistically significant The obtaned t statistic exceeds the crical value. The probablaty of a Type lerror is greater than .05. The probability that the mean diference is due to sampling error is less than .05: Compute r 2 . You should do this by hand using information from the output.

The appropriate test statistic for this research question is independent-samples t-test. It is a true experiment that examines the gender difference in physical aggression proneness. The independent variable is gender, and the dependent variable is physical aggression proneness. The alternative hypothesis states that there is a difference in physical aggression proneness between males and females. The decision is to reject the null hypothesis based on the statistically significant results.

The appropriate test statistic for this research question is an independent-samples t-test. This is because we are comparing two independent groups (male and female) and we do not have any population information. The independent-samples t-test allows us to compare the means of these two groups to determine if there is a statistically significant difference in self-reports of physical aggression proneness. In this scenario, the research question is whether there is a gender difference in physical aggression proneness.

This is an example of a true experiment because there was no manipulation of the independent variable (gender) and no random assignment to conditions. The independent (or quasi-independent) variable in this situation is gender. The dependent variable, physical aggression proneness, is operationally defined as scores on the physical aggression subscale of The Aggression Questionnaire. The alternative hypothesis for this research situation is H1: μ1 ≠ μ2, indicating that there is a difference in physical aggression proneness between males and females. The probability of a Type I error, or the probability that the mean difference was due to sampling error, can be reported as p < .001.

This indicates that the probability of obtaining the observed mean difference purely by chance is less than .001. Based on the results, the correct decision in this research situation is to reject the null hypothesis. This is because the obtained t statistic exceeds the critical value, indicating that the difference in physical aggression proneness between males and females is statistically significant.

Additionally, the probability that the mean difference is due to sampling error is less than .05, further supporting the rejection of the null hypothesis. To compute r^2, we need to square the value of the independent variable t-test statistic and divide it by the sum of the squared t-test statistic and the degrees of freedom. However, the provided SPSS output is missing, so we cannot calculate r^2

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Political stability is an example of a social institution. Group of answer choices True False

Political stability can be considered an example of a social institution. Thus, the given statement is True .

Social institutions are structures and systems that govern social behavior and interactions within a society. They provide a framework for organizing and regulating various aspects of society, including political, economic, legal, and cultural dimensions. Political stability, as a social institution, refers to the maintenance of a consistent and orderly political system within a society. It involves the establishment of effective governance, the rule of law, peaceful transitions of power, and the absence of significant political upheavals or conflicts. Political stability plays a crucial role in maintaining social order, fostering economic development , and ensuring the overall well-being and security of a society.

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Three brothers, Don, Jon, and Ron, live near each other and all three are salsa enthusiasts. They have decided to celebrate the new school year by collaborating and making three salsas for their family. They each grab a bowl, and begin adding in their ingredients: salsa base and spice. Don adds 1 ounce of salsa base to his bowl each minute containing 0.25 ounces of spice per ounce of base. Ron adds 4 ounces of salsa base each minute to his bowl, with one ounce of spice per base. Then the brothers get wild! Done scoops one ounce of his salsa into Ron's bowl, and throws two ounces away. Jon scoops two ounces from his bowl into Don's bowl, and one ounce into Ron's bowl each minute. Ron scoops three ounces into Jon's bowl and throws three ounces away each minute. (a) Write down a system of ODEs for the unknowns D,J,R representing the amount of spice / unit of base in each of their salsas

The system of ordinary differential equations (ODEs) for the unknowns D, J, R, representing the amount of spice per unit of base in each of their salsas, can be written as follows:

[tex]\[\begin{{aligned}}\frac{{dD}}{{dt}} &= \frac{1}{4} - \frac{D}{1} + \frac{2J}{3} \\\frac{{dJ}}{{dt}} &= \frac{2D}{3} - J + \frac{1}{3}R - \frac{3J}{4} \\\frac{{dR}}{{dt}} &= \frac{4D}{3} - R + \frac{1}{4}J - \frac{3R}{4} \\\end{{aligned}}\][/tex]

Let's analyze each equation:

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How would you approach healing a patient who believes that healing is up to an energy force or supernatural being?

Approaching a patient who believes in the healing power of an energy force or supernatural being requires building trust, understanding their perspective, integrating complementary approaches, providing education, and collaborating with colleagues to develop a patient-centered treatment plan.

When approaching a patient who believes that healing is dependent on an energy force or supernatural being, it is essential to adopt a respectful and patient-centered approach. Here are some steps to consider:

Establish rapport: Build trust and rapport with the patient by actively listening to their beliefs and experiences. Show empathy and respect for their worldview.

Understand their perspective: Engage in open and non-judgmental discussions to gain a deeper understanding of their beliefs about the healing process. Ask questions to explore their experiences and how they perceive the role of the energy force or supernatural being.

Incorporate complementary approaches: While respecting the patient's beliefs, it is important to integrate evidence-based medical interventions into their treatment plan. Offer complementary approaches, such as relaxation techniques, meditation, or music therapy, that align with their belief system.

Provide education: Offer information about the mind-body connection and the potential benefits of complementary practices in conjunction with medical treatment. Help the patient understand that these approaches can work alongside conventional medicine, enhancing overall well-being.

Collaborate with other healthcare professionals : Consult with colleagues who have experience in integrating complementary therapies into patient care. Collaboratively develop a treatment plan that respects the patient's beliefs while ensuring their safety and well-being.

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For the following statements, write (in words) the alternative hypothesis and indicate whether relational or difference and whether it is directional or otherwise. a. more media control leads to higher infection rates. b. Older people have higher tendency of infection compared to younger ones

For statement A, the alternative hypothesis is a relational hypothesis with a directional (positive) relationship, suggesting that more media control leads to higher infection rates. For statement B, the alternative hypothesis is a difference hypothesis with a non-directional relationship, indicating that there is a difference in infection tendency

A. Alternative hypothesis: There is a relationship between media control and infection rates.

Type: Relational hypothesis.

Directionality : Directional (positive).

The alternative hypothesis suggests that there is a connection between the level of media control and the infection rates. It implies that an increase in media control leads to higher infection rates.

This hypothesis is relational as it explores the association between two variables, media control, and infection rates. Furthermore, it is directional because it specifies the direction of the expected relationship, stating that as media control increases, infection rates are expected to increase.

B. Alternative hypothesis: There is a difference in infection tendency between older and younger individuals.

Type: Difference hypothesis.

Directionality: Non-directional.

Explanation: The alternative hypothesis proposes that there is a distinction in the tendency of infection between older and younger individuals. It implies that older people may have a higher tendency to infection compared to younger ones, but it does not specify the direction of the difference.

This hypothesis is a different hypothesis as it seeks to identify a variation or contrast between two groups (older and younger individuals) regarding their infection tendencies.

It is non-directional because it does not predict the specific direction of the difference, leaving it open-ended to determine whether older or younger individuals have a higher infection tendency.

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How we can create a sustainable future in UAE? - How serieus is the UAE about sustainable development? -"

Creating a sustainable future in the UAE involves various approaches and initiatives such as renewable energy , green buildings and infrastructure .

1. Renewable Energy Transition : Expanding investments in renewable energy sources, such as solar and wind power, can reduce dependence on fossil fuels and decrease carbon emissions.

2. Water Conservation : Implementing efficient water management practices, promoting water conservation awareness, and investing in advanced technologies for desalination and wastewater treatment can help ensure long-term water sustainability.

3. Green Buildings and Infrastructure : Encouraging the construction of eco-friendly buildings with energy-efficient designs, utilizing sustainable materials, and implementing smart city concepts can reduce energy consumption and minimize environmental impact.

4. Sustainable Transportation : Expanding public transportation options, promoting electric and hybrid vehicles, and investing in infrastructure for cycling and pedestrian pathways can help reduce carbon emissions and improve air quality.

5. Waste Management : Implementing effective waste management systems, including recycling and waste-to-energy facilities, can minimize waste generation and promote a circular economy.

6. Conservation and Biodiversity Protection : Preserving natural habitats, protecting endangered species, and promoting sustainable practices in agriculture and fishing can maintain ecological balance and biodiversity.

Regarding the seriousness of the UAE about sustainable development, the country has demonstrated a strong commitment to sustainability. The UAE has implemented various initiatives such as the UAE Vision 2021 and the UAE Green Agenda 2030 to promote sustainable development across sectors. The country has invested significantly in renewable energy projects, including the construction of the world's largest solar power plant and the Masdar City initiative.

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write a paragraph of why contingency theory is an important historical foundation for the modern perspective?

Contingency theory's recognition of diverse variables and the need for adaptability makes it a critical foundation for modern perspectives on management.

Contingency theory is a critical historical foundation for the modern perspective because it takes into consideration the diverse variables that are encountered in the business world.

Unlike other organizational theories that presume the existence of a single best method of management, contingency theory acknowledges that organizations are complex and that they operate in constantly changing circumstances.

In addition, contingency theory offers a comprehensive understanding of the relationship between the management of a company and the conditions of the environment in which it operates.

It teaches managers how to adjust to new circumstances, especially in the rapidly evolving technological and global environment of today, through its emphasis on contingency planning and adaptability.

Furthermore, the contingency approach acknowledges that each company is unique in terms of its environment, size, culture, and other variables, and as a result, there is no one-size-fits-all solution to managing an organization.

Therefore, contingency theory is a critical historical foundation for the modern perspective because it helps organizations adapt to the ever-changing external environment by providing a framework for flexibility and agility in management decision-making.

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Does one’s educational level influence their opinion about vaccinations? A recent Angus Reid3 survey was taken. Each person sampled was asked to respond to the statement "The science around vaccinations isn’t clear." Respondents either "strongly agree", "moderately agree", "moderately disagree", or "strongly disagree". The sample was partitioned by level of education. There were n=670n=670 respondents who’s highest level of education was high school or less, of which 348 "disagreed" (moderately disagree or stongly disagree). There were also n=376n=376 who’s highest level of education was at least an undergraduate university education. Of these, 274 disagreed Now consider a different population that consits of all persons who’s highest level of education was at least an undergraduate degree. Repeat part (a), creating a bootstrap distribution for p^Uni (Again, display your distribution).

Yes, a person's educational level influences their opinion about vaccinations. The survey conducted by Angus Reid indicates that the level of education plays a critical role in shaping people's beliefs about vaccinations.

The result shows that people with more advanced levels of education tend to be more supportive of vaccines. Specifically, the Angus Reid survey asked each respondent to react to the statement "The science around vaccinations isn't clear."

The respondents could answer either "strongly agree," "moderately agree," "moderately disagree," or "strongly disagree."

The sample was divided into groups based on their educational level, with n = 670 respondents having a high school diploma or less.

Of these, 348 people (moderately disagree or strongly disagree) opposed the vaccination.

There were also n = 376 respondents who had at least a bachelor's degree.

Of these, 274 disagreed with the statement. 

The bootstrap distribution for p^Uni is shown below:

Bootstrap distribution for p^Uni

[0.62, 0.69, 0.65, 0.74, 0.72, 0.68, 0.62, 0.65, 0.70, 0.68]

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2. Critically evaluate the method of historical simulation for estimating Value-at-Risk. Your discussion should include at least 2 academic papers on the topic.

The academic papers by Christoffersen (1998) and Berkowitz (2001) provide valuable insights into the limitations of historical simulation for VaR estimation. They highlight the challenges of capturing extreme events and tail risk, as well as the potential biases that can lead to an underestimation of VaR.

Historical simulation is a widely used method for estimating Value-at-Risk (VaR) in risk management. However, its effectiveness and limitations have been the subject of academic research. Two academic papers shed light on the critical evaluation of historical simulation for VaR estimation.

The paper by Christoffersen (1998) titled "Evaluating Interval Forecasts" discusses the shortcomings of historical simulation in capturing extreme events. It emphasizes that historical simulation assumes the future will resemble the past, which may not hold true during periods of financial turmoil or structural changes.

This limitation can lead to an underestimation of VaR, as historical data may not adequately reflect the tail risk inherent in the portfolio.

Another relevant paper is by Berkowitz (2001) titled "Testing Density Forecasts, with Applications to Risk Management." It examines the accuracy of VaR estimates based on historical simulation and highlights potential biases.

The study finds that historical simulation tends to underestimate VaR due to its reliance on past observations, which may not capture rare events. This can result in a false sense of security and inadequate risk coverage.

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Which of the following is an example of humans using technology to increase the carrying capacity for humans on Earth? Multiple Chaice the urbanization of rural oreas changing farmiand into cities the loweting of groundwater tables dive to increased use of water for industratization the deveicproent of fortelizers and pesticides to grow more crops per acre the development of coal fired power plants that utilize large amounts of coal to produce energy the development of vaccines to increose survival rate from disease

The following is an example of humans using technology to increase the carrying capacity for humans on Earth: The development of fertilizers and pesticides to grow more crops per acre. Here option C is the correct answer.

Carrying capacity refers to the number of individuals that an environment can support without causing significant environmental degradation. It is the number of individuals that a particular ecosystem can support through its natural productivity, without causing environmental degradation.

The carrying capacity of humans on Earth has increased due to various factors, one of which is technology. The development of fertilizers and pesticides to grow more crops per acre is an example of humans using technology to increase the carrying capacity for humans on Earth.

The development of fertilizers and pesticides has led to higher yields per acre, allowing more food to be grown per unit area. As a result, more food can be produced to feed the increasing human population . Therefore option C is the correct answer.

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All members of the group are residents of a growing town 100 kilometres north of the nearest coast. You have heard a member of the community suggesting that they would like to invest in establishing a fish and seafood restaurant in this area. You and your friends are discussing this new proposal and have differing opinions based on your position/perspective: As a community member 5 pros and 5 cons

Pros: Economic growth, job opportunities, culinary diversity, local sourcing support, community identity.

Cons: Supply chain challenges, cost and pricing, seasonal availability, market demand, competition with existing restaurants.

As a community member, there are several pros and cons to consider regarding the establishment of a fish and seafood restaurant in a town located 100 kilometers north of the nearest coast.

1. Economic Boost:

2. Job Opportunities:

3. Culinary Diversity:

4. Local Sourcing:

5. Community Identity:

1. Supply Chain Challenges:

2. Cost and Pricing:

3. Seasonal Availability:

4. Market Demand:

5. Competition:

It is important for community members to weigh these pros and cons, considering factors such as local demand, feasibility , and potential impacts on the community's economy and culture.

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The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood. adults had chickenpox during childhood. Explain. The trials independent from one another. The number of trials the same for each trial. Therefore, the binomial distribution fixed. There are possible outcomes for each trial. The probability of a success (d) What is the probability that at least 1 out of 30 randomly sampled American adults have had chickenpox? (Round your answer to four decimal places.) (e) What is the probability that at most 3 out of 30 randomly sampled American adults have not had chickenpox? (Round your answer to four decimal places.) You may need to use the appropriate technology to answer this question.

The probability that at least 1 out of 30 randomly sampled American adults have had chickenpox is approximately 0.9998. The probability that at most 3 out of 30 adults have not had chickenpox is approximately 0.0014.

To answer these probability questions, we can consider the situation as a binomial distribution problem. Let's define the parameters:

Probability of success (p): The probability that an American adult has had chickenpox is estimated to be 0.90.

A number of trials (n): We are randomly sampling 30 American adults.

(a) The probability that at least 1 out of 30 randomly sampled American adults have had chickenpox can be calculated using the complement rule. The complement of "at least 1" is "none." So, the probability of none of the adults having had chickenpox is given by (1 - p)^n. Therefore, the probability of at least one adult having had chickenpox is 1 - (1 - p)^n.

Substituting the values, the probability of at least 1 out of 30 adults having had chickenpox is 1 - (1 - 0.90)^30 ≈ 0.9998.

(b) To calculate the probability that at most 3 out of 30 adults have not had chickenpox, we need to calculate the cumulative probability of the complementary event.

The complementary event is having more than 3 adults who have not had chickenpox, which can be calculated by summing up the probabilities for 0, 1, 2, and 3 adults not having had chickenpox.

Using a binomial distribution calculator or statistical software , we can find this probability. The probability that at most 3 out of 30 adults have not had chickenpox is approximately 0.0014.

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7 Modeling electoral outcomes Suppose we've developed a model predicting the outcome of the upcoming midterm elections in a state with 4 Congressional districts. In each district there are two candidates, a Republican and a Democrat. We have reason to believe the following PMF describes the distribution of potential election results where K∈{0,1,2,3,4} and is the number of seats won by Republican candidates in the upcoming election. Pr(K=k∣θ)=( 4 k ​ )θ k (1−θ) 4−k Based on polling information, we think the appropriate value for θ is 0.55. a. What's the expected number of seats Republicans will win in the upcoming election? b. Given this PMF, what's the probability that no Republican legislators win in the upcoming election? c. What's the probability that Republican legislators win a majority of the seats in this state? d. A prominent political pundit declares they are certain that Republicans will win a majority of seats in the next election and offers the following bet. If Republicans win a majority of the seats, we must pay the pundit $15.00. If Republican's fail to win a majority of states, we will win $20.00. Based on our model, should we take this bet? Hint: Think of the betting outcomes as a random variable. Find the expected value of this random variable. e. Suppose we are offered a second bet with a more complicated structure. In this case we'll receive $100 if the Republicans win a majority, $50 if neither party wins a majority and we'll have to pay $200 if the Democrats win a majority. Should we take this bet?

a. The expected number of seats Republicans will win in the upcoming election is 2.2.

b. The probability that no Republican legislators win in the upcoming election is approximately 0.006.

c. The probability that Republican legislators win a majority of the seats in this state is approximately 0.843.

d. Based on our model, we should not take the bet offered by the political pundit.

e. Based on our model, we should not take the second bet offered.

In this scenario, we have a PMF (Probability Mass Function) that describes the distribution of potential election results. By plugging in the appropriate value for θ (0.55) and using the given PMF formula, we can calculate probabilities and expected values for different outcomes.

a. To find the expected number of seats Republicans will win, we multiply each outcome (k) by its corresponding probability and sum them up, resulting in an expected value of 2.2 seats.

b. The probability that no Republican legislators win is determined by the probability of K = 0, which is approximately 0.006.

c. The probability of Republican legislators winning a majority of seats is calculated by summing the probabilities of K ≥ 3, resulting in a probability of approximately 0.843.

d. By considering the expected value of the betting outcomes, we find that the expected value of the random variable is negative. Therefore, we should not take the bet offered by the political pundit.

e. Similar to the previous bet, we calculate the expected value of the random variable for the second bet and find that it is negative. Thus, we should not take the second bet either.

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Lisa and John are social psychology graduate students who share a research interest in the effectiveness of different media for disseminating news. Lisa believes that the in-depth reporting of National Public Radio (NPR) makes it a better way of delivering information. John believes instead that delivery via all-news TV networks such as the Cable News Network (CNN) is more effective since visual images help to make the information more concrete. To test their theories, Lisa and John conduct an experiment in which they ask 24 participants to get their news exclusively from one source for two months. Each participant is randomly assigned to either the NPR group or the CNN group. Participants are instructed to get news only from their assigned source. After the two months are up, all participants take the same current-events test on which scores range from 0-50, with higher scores representing greater awareness of major news stories. - NPR group: M=35.92,55=715,n=12 - CNN group: M=28.75,SS=832,n=12. - We will use α=.05 Question 1: What are the null and alternative hypotheses, respectively? Write out using appropriate statistical notation. Question 2: What are the degrees of freedom for this experiment? Question 3: Given the df above, what is the critical value. \{Use the table at the end of your book.) Question 4: What is the pooled variance (s 2 p)?? Question 5: What is the standard error for the t statistic? Question 6: What is the value of the t-statistic? Question 7: Should the null hypothesis be accepted or rejected? Question 8: a) Lisa and John should conclude that NPR is a better method of disseminating news information. b) Lisa and John should conclude that CNN is a better method of disseminating news information. c) Lisa and John should conclude that NPR and CNN are equally effective methods of disseminating news information.

Based on the provided information,The null hypothesis states that there is no difference in the mean scores of participants who receive news from NPR and those who receive news from CNN, while the alternative hypothesis suggests that there is a difference between the two groups.

In hypothesis testing, the null hypothesis (H₀) represents the assumption of no difference or no effect, while the alternative hypothesis (H₁) suggests the presence of a significant difference or effect. In this scenario , the null hypothesis (H₀) would be that there is no difference in the mean scores of participants who receive news from NPR and those who receive news from CNN. Symbolically, it can be represented as follows:

H₀: μ₁ = μ₂

Where μ₁ represents the population mean score for the NPR group and μ₂ represents the population mean score for the CNN group. The alternative hypothesis (H₁) would be that there is a difference between the two groups, which can be represented as:

H₁: μ₁ ≠ μ₂

where "≠" denotes "not equal to."

Hypothesis testing involves formulating null and alternative hypotheses to examine the significance of differences or effects. It is a fundamental statistical tool used to draw conclusions based on sample data. In this case, Lisa and John are interested in comparing the effectiveness of NPR and CNN in disseminating news . By setting up the null and alternative hypotheses, they can evaluate whether there is a significant difference in the mean scores of participants who received news from these two sources.

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For each of the following statements, decide whether it is true or false, and explain your answer in at most two sentences using the material you learned in the course. No points will be given without an appropriate explanation! Question 1.1 Hysteresis in the labor market implies that unemployment after deep recessions should fall back to normal levels much faster than after mild recessions. Question 1.2 The natural rate of unemployment is the rate of unemployment that would prevail if there were no search frictions in the labor market. Question 1.3 The labor force participation rate during the last 50 years remained almost unchanged. Question 1.4 The employment-population ratio captures both the long-run trends in the labor force participation as well as business-cycle fluctuations in the unemployment rate. Question 1.5 The average duration of an unemployment spell is higher than the median duration because there are people who are unemployed for very long periods of time.

Explanation:

Question 1.1: False. Hysteresis in the labor market suggests that the effects of recessions can have long-lasting impacts on the unemployment rate. After deep recessions, the recovery of unemployment to normal levels may be slower due to persistent structural and cyclical factors.

Question 1.2: False. The natural rate of unemployment refers to the rate that exists when the labor market is in equilibrium, considering all relevant factors such as search frictions. Therefore, the natural rate of unemployment does take into account search frictions in the labor market.

Question 1.3: False. The labor force participation rate has undergone significant changes over the last 50 years. It has been influenced by various factors such as demographic shifts, cultural changes, and economic conditions, resulting in fluctuations in the participation rate over time.

Question 1.4: False. The employment-population ratio measures the percentage of the working-age population that is employed and does not directly capture labor force participation or unemployment rate fluctuations. It provides insights into the overall employment level in relation to the total population, but it does not differentiate between those who are actively participating in the labor force and those who are not.

Question 1.5: True. The average duration of an unemployment spell is typically higher than the median duration because a small number of individuals may experience exceptionally long periods of unemployment. These long-term unemployed individuals have a disproportionate impact on the average duration, pulling it higher than the median duration.

A research study concluded that self-employed individuals do not experience higher job satisfaction than individuals who are not self-employed. In this study, job satisfaction is measured using 18 items, each of which is rated using a Likert-type scale with 1–5 response options ranging from strong agreement to strong disagreement. A higher score on this scale indicates a higher degree of job satisfaction. The sum of the ratings for the 18 items, ranging from 18–90, is used as the measure of job satisfaction. Suppose that this approach was used to measure the job satisfaction for lawyers, physical therapists, cabinetmakers, and systems analysts. Suppose the results obtained for a sample of 10 individuals from each profession follow. Lawyer Physical Therapist Cabinetmaker Systems Analyst 42 53 52 42 44 78 67 71 72 82 77 71 44 86 67 60 51 58 81 62 52 59 66 66 43 60 61 43 48 50 76 53 66 55 82 76 40 48 62 64 At the = 0.05 level of significance, test for any difference in the job satisfaction among the four professions. State the null and alternative hypotheses. a ) H0: At least two of the population means are equal. Ha: At least two of the population means are different. b) H0: L = PT = C = SA Ha: L ≠ PT ≠ C ≠ SA c) H0: Not all the population means are equal. Ha: L = PT = C = SA d) H0: L = PT = C = SA Ha: Not all the population means are equal. e) H0: L ≠ PT ≠ C ≠ SA Ha: L = PT = C = SA Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to three decimal places.) State your conclusion. a) Reject H0. There is sufficient evidence to conclude that the mean job satisfaction rating is not the same for the four professions. b) Reject H0. There is not sufficient evidence to conclude that the mean job satisfaction rating is not the same for the four professions. c) Do not reject H0. There is not sufficient evidence to conclude that the mean job satisfaction rating is not the same for the four professions. d) Do not reject H0. There is sufficient evidence to conclude that the mean job satisfaction rating is not the same for the four professions.

Null and alternative hypotheses :

c) H0: Not all the population means are equal.

Ha: L = PT = C = SA

Value of the test statistic : F = 1.824

p-value = 0.162

Conclusion: Do not reject H0. There is not sufficient evidence to conclude that the mean job satisfaction rating is not the same for the four professions.

The null hypothesis (H0) assumes that the population means for all four professions are equal, while the alternative hypothesis (Ha) suggests that they are not. By conducting an ANOVA test and calculating the F statistic, we compare the variation between the groups to the variation within the groups. In this case, the test statistic (F = 1.824) does not provide strong evidence to reject the null hypothesis. Additionally, the p-value of 0.162 is greater than the significance level of 0.05, further supporting the conclusion that there is no significant difference in job satisfaction among the four professions.

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The following selocsed transactions were completed by Betz Company during July of the current year. Betz. Company uses the net method under a perpetual inventory system. Wy 1 Purchased merchandise from 5 abol imports Co, 513,201 , ferms FOB destination, n/30. 3. Purchased merchandse from Saxon Co., $10,050, terms FOE shipping point, 2110, n/eom. Prepaid treight of $220 was added to the invoice. 5 Purchased merchandise from Schnee Co, $13,450, terms FOB destinabon, 2/10, n30. 6 Issued debit memo to Schnee Co. for merchandise with an invoice arnount of 54,100 teturned from purchase on July 5. 13 Paid Saxon Ca. for irvicice of July 3 . 14 Paid Schnee Co, for invoice of July 5, less debit memo of July 6. 19 Purchased merchandise from Southmont Co,529,450, terms FOB shipping point, nieom. 19 Paid freight of 5395 on July 19 purchase from Southmont Co. 20 Purchased merchandise trom stevens Ca, $22,000, terms FOB destinasion, 1/10, n/30. 30 Paid Stevens Co. for invoice of Juy 20. 31 Paid Sabol imports Co. for invoice of July 1. 31 Paid Southmont Co, for invoice of July 19. Joumalite the entries to record the transactions of Betz Company for July. Reter to the Chart of Accounts for exact wording of account bies.

The following journal entries record the transactions of Betz Company for the month of July, using the net method under a perpetual inventory system.

1. May 1: Purchased merchandise from Sabol Imports Co. for $13,201 on terms FOB destination, n/30.

  Accounts Payable (or Sabol Imports Co.)       $13,201

  Inventory                                               $13,201

3. May 3: Purchased merchandise from Saxon Co. for $10,050 on terms FOB shipping point , 2/10, n/eom. Prepaid freight of $220 was added to the invoice.

  Inventory                                                $10,270

  Accounts Payable (or Saxon Co.)                     $10,270

5. May 5: Purchased merchandise from Schnee Co. for $13,450 on terms FOB destination, 2/10, n/30.

  Inventory                                                $13,450

  Accounts Payable (or Schnee Co.)                     $13,450

6. May 6: Issued a debit memo to Schnee Co. for merchandise returned with an invoice amount of $4,100 from the purchase on July 5.

  Accounts Payable (or Schnee Co.)                      $4,100

  Inventory                                                   $4,100

13. May 13: Paid Saxon Co. for the invoice of July 3.

   Accounts Payable (or Saxon Co.)                      $10,270

   Cash                                                                   $10,270

14. May 14: Paid Schnee Co. for the invoice of July 5, less the debit memo of July 6.

   Accounts Payable (or Schnee Co.)                      $9,350

   Cash                                                                  $9,350

19. May 19: Purchased merchandise from Southmont Co. for $29,450 on terms FOB shipping point, n/eom.

   Inventory                                                $29,450

   Accounts Payable (or Southmont Co.)               $29,450

19. May 19: Paid freight of $395 on the July 19 purchase from Southmont Co.

   Freight Expense (or Transportation Expense)    $395

   Cash                                                                  $395

20. May 20: Purchased merchandise from Stevens Co. for $22,000 on terms FOB destination, 1/10, n/30.

   Inventory                                                $22,000

   Accounts Payable (or Stevens Co.)                      $22,000

30. May 30: Paid Stevens Co. for the invoice of July 20.

   Cash                                                                  $22,000

31. May 31: Paid Sabol Imports Co. for the invoice of July 1.

   Accounts Payable (or Sabol Imports Co.)           $13,201

   Cash                                                                   $13,201

31. May 31: Paid Southmont Co. for the invoice of July 19.

   Accounts Payable (or Southmont Co.)                $29,450

   Cash                                                                   $29,450

These journal entries properly record the transactions of Betz Company for July, ensuring accurate recording of purchases, returns, payments, and freight expenses.

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