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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You wish to evaluate the differences in educational quality between two training centres (A and B). You take a random sample of 98 individuals trained at centre A, and an independent random sample of 41 individuals trained at centre B. The individuals in both samples take the same, standardised test. The mean test score of individuals in sample A is 85.1 and the sample standard deviation is 12.2. The mean test score in sample B is 67.4 and the sample standard deviation is 11.0. Find the lower limit of a 98% confidence interval for the difference between the population means. Form the confidence interval around a positive point estimate for the difference between the means. In order to obtain a positive point estimate, make sure you subtract the lower value from the higher value when calculating the difference between the sample means. (Provide your answer as a number rounded to 2 decimal places.)
You would like to estimate the variance and standard deviation for credit scores based on FICO. These scores are thought to be normally distributed. You randomly select 36 credit scores to get a standard deviation of 56.5. Using a 95% confidence interval, estimate the variance and standard deviation of credit scores.
The dean randomly selected 33 students who took the calculus final exam and looked at their scores on it. In this sample the mean was 80.3 and the standard deviation was 3.7. No information about the population's distribution is available. Construct a 98% confidence interval for the mean score of all students who took this exam. Round your answers to 2 places after the decimal point, if necessary. If it is not appropriate to construct a confidence interval in this situation, then enter "0" in both answer boxes below. 98% confidence interval:
Suppose that you randomly survey death records for people born in 1900 in Louisiana and compared the life span of two different ethnicities. Of the 109 individuals sampled from ethnicity 1, the mean life span was 45.5 years with a standard deviation of 18.2 years. Of the 105 individuals sampled from ethnicity 2, the mean life span was 44.9 years with a standard deviation of 11.4 years. Conduct a hypothesis test with a 10% level of significance to see if the mean life spans in Louisiana were the same for the two ethnicities. Step 1: State the null and alternative hypotheses. Ho: µ1 – µ2 =v Ha:H1 – µ2 # v (So we will be performing a two-tailed ♥ test.)
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?
Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 37 randomly selected people who train in groups, and finds that they run a mean of 47.7 miles per week. Assume that the population standard deviation for group runners is known to be 3.3 miles per week. She also interviews a random sample of 49 people who train on their own and finds that they run a mean of 49.4 miles per week. Assume that the population standard deviation for people who run by themselves is 4.4 miles per week. Test the claim at the 0.10 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
Researchers are interested in the effect of pet companionship on depression in elderly adults. They randomly select 25 participants from the target population and record their depression scores before and after spending a month with a dog (a higher score indicates a higher level of depression). The researchers then compute the difference scores for each participant (difference after before). The mean of differences is -1 and the standard deviation of differences is 2. Can the researchers conclude that there is an effect of pet companionship on depression in elderly adults? Conduct the appropriate statistical test (two-tailed) with a = .05. Should you reject or fail to reject the null hypothesis? Reject O Fail to reject
A study of 96 bolts of carpet showed that their average length was 77.5 yards. The standard deviation of the population is 1.6 yards. Which of the following is the 90% confidence interval for the mean length per bolt of carpet?
Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
A reading class can read an mean of 175 words per minute and a population standard deviation of 20 words per minutes. The top 3% of the class is to receive a special award for a random sample of 49 students. What is the minimum average number of words per minute a student would need to read in order to get the award for a random sample of 49 students? Assume the data is normally distributed.
Forty refrigerators from two different models of a refrigerator manufacturer are compared to see if the smaller model uses less energy than the larger model. The energy was measured in kilowatt hours per month (kwh/mo). From years of testing these models, it is known that the standard deviation of kwh/mo is 34 kwh/mo for the smaller model and 40 kwh/mo for the larger one. From this survey, the sample means and standard deviations are 125 and 34, and 90 and 40, respectively. If we conduct a hypothesis test, the decision and conclusion are Group of answer choices Do not reject the null hypothesis and conclude that the average kwh/mo are the same. Reject the null hypothesis and conclude that the average kwh/mo are different. Reject the null hypothesis and conclude that the average kwh/mo for the smaller refrigerator is higher than for the larger model. Do not reject the null hypothesis and conclude that the average kwh/mo for the smaller refrigerator is higher than for…

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