The text on the image outlines steps for using Student's t Distribution:### Student's t Distribution**Step 1**: Enter the number of degrees of freedom.- [Input box for degrees of freedom]**Step 2**: Select one-tailed or two-tailed.- ( ) One-tailed- (o) Two-tailed**Step 3**: Enter the test statistic. (Round to 3 decimal places.)- [Input box for test statistic]**Step 4**: Shade the area represented by the p-value.### Graph DescriptionOn the right side, there is a graph with a vertical axis labeled from 0 to 0.4, with intervals of 0.1. This graph is likely used to visualize the t distribution and the corresponding p-value as shaded areas under the curve. The graph is intended for plotting the calculated p-value based on the entered test statistic and the degrees of freedom. **Text Transcription and Explanation:**A very large company is interested in its employees' productivity. The company reports from its historical data that its employees spend a mean of 164 minutes per employee (on a typical day) dealing with email. To test this claim, an independent consultant chooses 26 employees at random and finds that those employees spend a sample mean of 172 minutes dealing with email, with a sample standard deviation of 21 minutes. Assume that the population of amounts of time employees spend dealing with email is approximately normally distributed.Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to reject the claim that μ, the mean number of minutes employees spend dealing with email, is equal to 164.(a) State the null hypothesis $ H_0 $ and the alternative hypothesis $ H_1 $ that you would use for the test.Diagram or Graph Explanation:The diagram contains a box with symbols representing different statistical symbols and operators used to state the null and alternative hypotheses:- $ \mu $ represents the population mean.- $ \bar{x} $ is the sample mean.- The symbols include equality ($ = $), not equal ($ \neq $), and inequality signs ($ <, >, \leq, \geq $).(b) Perform a t test and find the p-value.Here is some information to help you with your t test.(The image does not contain further details for performing the t test or give explicit values for the p-value, so this part is left open-ended for educational purposes.)**Notes:**- The company claims that the mean is 164 minutes.- Sample statistics: mean = 172 minutes, standard deviation = 21 minutes, sample size = 26.- The significance level is 0.05.- Hypotheses can be framed as: - Null hypothesis $ H_0: \mu = 164 $ - Alternative hypothesis $ H_1: \mu \neq 164 $

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Answered: A very large company is interested in…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You may need to use the appropriate technology to answer this question. Barron's has collected data on the top 1,000 financial advisers. Company A and Company B have many of their advisers on this list. A sample of 16 of the Company A advisers and 10 of the Company B advisers showed that the advisers managed many very large accounts with a large variance in the total amount of funds managed. The standard deviation of the amount managed by the Company A advisers was s1 = $582 million. The standard deviation of the amount managed by the Company B advisers was s2 = $484 million. Conduct a hypothesis test at ? = 0.10 to determine if there is a significant difference in the population variances for the amounts managed by the two companies. What is your conclusion about the variability in the amount of funds managed by advisers from the two firms? State the null and alternative hypotheses. H0: ?12 ≠ ?22 Ha: ?12 = ?22 H0: ?12 ≤ ?22 Ha: ?12 > ?22 H0: ?12 = ?22 Ha: ?12 ≠ ?22 H0:…
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Students preparing for comprehensive exams usually spend many hours studying. Suppose it is known that the time students spend studying for comprehensive exams has a distribution that is skewed heavily to the right with a mean of 41.3 hours and a standard deviation of 6.8 hours. If a simple random sample of 72 students is selected and the amount of time each spent studying for the comprehensive exam is determined. State and check the two assumptions.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂
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light examined data on employment and answered questions regarding why workers separate from their employes. According to the article, the standard deviation of the length of time that women with one job are employed during the first 8 years of their career is 92 weeks. Length of time employed during the first 8 years of career is a left skewed variable. For that variable, do the following tasks. A. determine the sampling distribution of the sample mean for simple random samples of 50 women with one job. Explain your reasoning B. Obtain the probability that the sampling error made in estimating the mean length of time employed by all women with one job by that of a random sample of 50 such women will be at most 20 weeks
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A community college Math instructor feels that the average age of community college students decreased since they started teaching 25 years ago. Looking up data from the year the instructor started teaching, the mean age of community college students was 24.3 years old, with a population standard deviation of 2.2 years. The Math instructor took a sample of 40 of their current community college students and found that the sample had a mean age of 23.7 years old. Assume the population of ages of community college students is normally distributed and that the population standard deviation has remained the same. Test at the 5% significance level if the mean age of community college students is now less than 24.3 years old. The question is attached.
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