Do people behave differently on Friday the 13th? Researchers collected data on the number of shoppers at a random sample of45 grocery stores on Friday the 6th and Friday the 13th in the same month. Then they calculated the difference (6th – 13th) inthe number of shoppers at each store on these 2 days. The mean difference is –45.6 and the standard deviation of the differencesis 178.Construct and interpret a 90% confidence interval for the true mean difference.178O -45.6 + 1.645-V45We are 90% confident that this interval captures the true mean difference (6th – 13th) in the number of shoppers atgrocery stores.178O -45.6 ± 1.645-V44We are 90% confident that this interval captures the true mean difference (6th – 13th) in the number of shoppers atgrocery stores.178O -45.6 + 1.684-V45We are 90% confident that this interval captures the true mean difference (6th – 13th) in the number of shoppers atgrocery stores.178-45.6 + 1.684-V44We are 90% confident that this interval captures the true mean difference (6th – 13th) in the number of shoppers atgrocery stores.

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Answered: Do people behave differently on Friday…

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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A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 88.1. He knows the population standard deviation for the evening classes to be 4.2 points. A random sample of 150 students from morning classes results in a mean test score of 89.1. He knows the population standard deviation for the morning classes to be 7.7 points. Test his claim with a 95 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.
I need help, I keep getting it wrong and only have one more try for each.
Suppose the mean height in inches of all 9th grade students at one high school is estimated. The population standard deviation is 5 inches. The heights of 9 randomly selected students are 73, 66, 64, 67, 70, 63, 64, 72 and 72. = _______ Margin of error at 90% confidence level = ________ 90% confidence interval = _______, _________
An agriculture researcher plants twenty five plots with a new variety of corn. The average yielded for these plots is x-bar = 150 bushels per acre. Assume that the yield per acre for the new variety of corn follows a normal distribution with unknown mean and standard deviation = 10 bushels per acre. A 90% Confidence interval for mean is?
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1__μ2
From a random sample of 64 high school seniors, the sample mean and standard deviation of the math scores are found to be 90 and 7 respectively. Determine a 92% confidence interval for the mean math score of all seniors in the school and interpret your results. Please help me. Thank you
You would like to estimate the variance of sick days at a Fortune 500 company. You randomly select 15 employees and record how many sick days they have used in the past year and you get a sample standard deviation of 1.2. Provide a 95% confidence interval estimate of the population variance.
Please answer all. Thank you. A national random sample of 20 NCAE Scores from 2010 is listed below. 29,26,13,23,23,25,17,22,17,19 12,26,30,30,18,14,12,26,17,18 Calculate the 95 percent confidence interval for the mean NCAE score if the sample standard deviation is about 5.9407. A.) between 18.22 and 23.48 B.) between 18.07 and 23.63 C.) between negative infinity and 23.15 D.) between negative infinity and 23.05 The value of z selected for constructing a given confidence interval is called A.) confidence B.) error C.) student D.) critical value
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Over the past 10 years, contestants at the county fair hotdog eating contest averaged 125 hotdogs within the 1 minute time frame. (How did they do this? I have no idea). Anyway, due to Covid19 restrictions, while we continue to assume everything is normal, including distributions, there were only 15 contestants who ate an average of 122 hotdogs. s=6.5 and amazingly enough, everyone's hotdog quantity eaten was within a very narrow quantity difference. We're interested in finding out if there is significant change in hotdog eating. Use the hypothesis that H0: µ=125, Ha: #125. What distribution should we use? binomial distribution unimodal distribution z-distribution t-distribution
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 85.6. He knows the population standard deviation for the evening classes to be 4.6 points. A random sample of 150 students from morning classes results in a mean test score of 86.7. He knows the population standard deviation for the morning classes to be 8.3 points. Test his claim with a 99 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 3 of 3: Draw a conclusion and interpret the decision. Answer 围 Tables E Keypad Keyboard Shortcuts We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the…

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