A savings and loan association needsinformation concerning the checking accountbalances of its local customers. A randomsample of 14 accounts was checked and yieldeda mean balance of $664.14 and a standarddeviation of $ 297.29. Find a 90% confidenceinterval for the true mean checking accountbalance for local customers.O a.O b.O c.O d.523.43 to 804.85493.71 to 834.57492.52 to 835.76455.65 to 872.63

Given datan=14x¯=664.14s=297.29confidence interval is 90%

Answered: A savings and loan association needs…

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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Given the following: Sample Mean = 84.3 Population Variance = 144 Sample size = 35 a. Calculate a 90% confidence interval for the mean. Interpret this interval. b. Calculate a 99% confidence interval for the mean. c. Explain the relationship between Confidence Levels and the resulting intervals.
The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689 find a 95% confidence interval for the standard deviation of the annual medical cost for all households. Assume medical costs are normally distributed.
A study reported that finger rings increase the growth of bacteria on health-care workers’ hands. Research suggests that 31 percent of health-care workers who wear rings have bacteria on one or both hands. Suppose that independent random samples of 100 health-care workers wearing rings is selected. What is the standard deviation of the sampling distribution of the sample proportions of health-care workers having bacteria on one or both hands? A. 100 B. 0.0462 C. 0.69
A consumer affairs investigator records the repair cost for 4 randomly selected TVs. A sample mean of $75.89 and standard deviation of $13.53 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Construct the 80% confidence interval. Round your answer to two decimal places.
The monthly earnings of a group of business students are normally distributed with a standard deviation of 478 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 91% and a margin of error of 125 dollars. O a. 56 O b. 43 O c. 46 O d. 64
The management of a national automobile manufacturer advertises a free transmission if it fails. The management expects transmissions will be normally distributed with a mean of 65,000 miles and a standard deviation of 2,146 miles. If management wants to give free transmission replacements to at most one percent of its automobile purchasers, at what mileage should warranty state replacement?
It is believed that nearsightedness affects 14% of children in a certain region. A school district tests the vision of 180 incoming kindergarten children. How many would be expected to be nearsighted? What is the standard deviation for the number of nearsighted children in such a group? You may assume that the children are independent of each other. The mean is? and the standard deviation is?
A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of S 297.29. Find a 90% confidence interval for the true mean checking account balance for local customers. Select one: a. 492.52 to 835.76 b. 455-65 to 872.63 C. 493-71 to 834-57 d. 523.43 to 804.85
MC Qu. 10-64 Sales at a fast-food restaurant average... Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to Increase daily sales. In order to determine the effectiveness of the advertising campaign, a sample of 49 day's sales were taken. The sample showed average daily sales of $6,300. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? Multiple Choice Fail to reject the null hypothesis. Reject the null hypothesis and conclude the mean is higher than $6,000 per day. Reject the null hypothesis and conclude the mean is lower than $6,000 per day. Reject the null hypothesis and conclude that mean equal to $6,000 per day.
?
The manager of a Kroger grocery store claims a membership card will save consumers more than $15 per week on average. To check this claim, a random sample of 12 shoppers with membership cards was obtained and their weekly grocery bills were inspected. Suppose the sample mean was $16.35 and the sample standard deviation was $3.20. Assume the distribution of savings is normal. a. Check the necessary assumptions and conditions for doing this test. b. What are the appropriate hypotheses to test the claim? c. Calculate the appropriate test statistic and p-value. d. What is the conclusion?
A food distribution company claims that a restaurant chain receives, on average, 26 pounds of fresh vegetables on a daily basis. The standard deviation of these shipments is known to be 4.4 pounds. The district manager of the restaurant chain decides to randomly sample 35 shipments from the company and finds a mean weight of 24.7 pounds. Test at a 3% level of significance to determine whether or not the food distribution company sends less than 26 pounds of fresh vegetables. a. Check the TWO requirements that are satisfied. The Central Limit Theorem applies. The a distribution is normal since n > 30. The a distribution is normal since the x distribution is normal. The p distribution is normal since np > 5 and nq > 5.

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