A marketing consultant observes 50 consecutive shoppers at a supermarket, recording how much each shopper spends in the store. What can be said about using these data to carry out a significance test about the mean amount spent by all shoppers at this supermarket? It would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at the supermarket because this sample wasn't randomly selected. It would be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at the supermarket because the supermarket could really benefit from the data. It would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at the supermarket because the sample isn't big enough. O It would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at the supermarket unless the supermarket had at least 500 shoppers in it.

Customers of a restaurant have been complaining about food poisoning over the past week. A health professional was called to investigate. He tested a representative sample of the eggs and reported that  23%  of the sample eggs tested positive for the salmonella bacterium. If a second health professional tests 4 eggs at random, what is the mean of the number of eggs that test positive for salmonella?

An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 240 Fundash runners and 200 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 240 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.2 minutes. The 200 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.4 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, μ1 , of Fundash is not equal to the mean completion time, μ2 , of Coolsprint?…

The personality characteristics of business leaders (e.g., CEOS) are related to the operations of the businesses that they lead (Oreg & Berson, 2018). Traits like openness to experience are related to positive financial outcomes and other traits are related to negative financial outcomes for their businesses. Suppose that a board of directors is interested in evaluating the personality of their leadership. Among a sample of n = 16 managers, the sample mean of the openness to experiences dimension of personality was M = 4.50. Assuming that u = 4.24 and o = 1.05 (Cobb-Clark & Schurer, 2012), use a two-tailed hypothesis test with a = .05 to test the hypothesis that this company's business leaders' openness to experience is different from the population. Standard Normal Distribution Mean - 0.0 Standard Deviation 1.0 .7198 .1401 .1401 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 -1.08 1.08 Step 1. Ho: ; H;: a = .05. Step 2. The critical region consists of Step 3. For these data the standard error is and…

A researcher wanted to find the best viewing distance for watching a TV screen. He recruited the following number of participants and asked them to evaluate their satisfaction of the viewing distance. Then, he conducted an ANOVA to see if there is any significant mean differences. Here are the number of participants by each group: Group 1 (4 feet) - 5 people Group 2 (7 feet) - 5 people Group 3 (10 feet) - 6 people Group 4 (13 feet) - 5 people Group 5 (16 feet) - 4 people Given the numbers as presented above, what is the degrees of freedom for within groups for this analysis? a. 4 b. 20 O c. 21 d. 25

A nationwide job recruiting firm wants to compare the annual incomes for childcare workers in Connecticut and California. Due to recent trends in the childcare industry, the firm suspects that the that the mean annual income of childcare workers in the state of Connecticut is greater than the mean annual income of childcare workers in California. To see if this is true, the firm selected a random sample of 10 childcare workers from Connecticut and an independent random sample of 10 childcare workers from California and asked them to report their mean annual income. The data obtained were as follows. 80°F Partly sunny Annual income in dollars Connecticut 42438, 52287, 46837, 49334, 52848, 52929, 57111, 51860, 44967, 52174 California 50998, 38334, 56804, 50919, 49446, 41141, 45717, 43963, 55156, 47295 LL D 144Hz Send data to calculator V The population standard deviations for the annual incomes of childcare workers in Connecticut and in California are estimated as 6300 and 6100,…

At a local food processing plant, it used to take an average of 90 minutes for new workers to learn a food processing job. Recently the company installed a new food processing machine. The supervisor at the company wants to find if the mean time taken by new workers to learn the food processing procedure on this new machine is different from 90 minutes. A sample of 20 workers showed that it took, on average, 85 minutes for them to learn the food processing procedure on the new machine. It is known that the learning times for all new workers are normally distributed with a population standard deviation of 7 minutes. Use a significance level of 0.01 to test the claim that the mean time is different from 90 minutes. What is the decision?   A) Reject Ho because the test statistic is in the critical region and the p-value is greater than the significance level. B) Reject Ho because the test statistic is not in the critical region and the p-value is greater than the significance level.…

At a local food processing plant, it used to take an average of 90 minutes for new workers to learn a food processing job. Recently the company installed a new food processing machine. The supervisor at the company wants to find if the mean time taken by new workers to learn the food processing procedure on this new machine is different from 90 minutes. A sample of 20 workers showed that it took, on average, 85 minutes for them to learn the food processing procedure on the new machine. It is known that the learning times for all new workers are normally distributed with a population standard deviation of 7 minutes. Use a significance level of 0.01 to test the claim that the mean time is different from 90 minutes. What is the test statistic?   A) -7.01 B) -2.539 C) -3.19 D) -2.528 E) -1.96