b. It has been reported that the average credit card debt for the newly employed is$3350. The hiring agency feels that their employees have a debt much less thanthis, so it conducts a study on 50 randomly selected new employees and findsthat the average debt is $3050. Given the population standard deviation is $1000.i. Set up the hypothesis test.At the 0.025 significance level is there enough evidence to suggest theaverage debt has changed?ii.

Answered: Image /qna-images/answer/67abfa37-e025-4640-93a3-24bd1ee997a6.jpg

Answered: b. It has been reported that the…

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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An auditor, pressed for time and on a low budget, is concerned with account errors on customer credit. She takes a sample of 25 accounts, and finds the account error, on average, is $-23, with a standard deviation of $16. Is the claim of the manager that errors average only $-10 supported? In this case, the null hypothesis is what?
The average starting salary of students who graduated from colleges of Business in 2009 was $48,800. A sample of 100 graduates of 2010 showed an average starting salary of $50,000. Assume the standard deviation of the population is known to be $7000. We want to determine whether or not there has been a significant increase in the starting salaries. a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested at the 5% level of significance. Determine the critical value for this test. d. What do you conclude from the test? e. What does the test result mean in the context of this problem?
In a survey, 26 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $40 and standard deviation of $6. Find the margin of error at a 90% confidence level.
In a survey, 41 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $36 and standard deviation of $4. Find the margin of error for a 95% confidence level.
Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to increase daily sales. To determine the effectiveness of the advertising campaign, a sample of 49 days of sales were taken. They found that the average daily sales were $6,300 per day. From past history, the restaurant knew that its population standard deviation is about $1,000. The restaurant wishes to test whether sales have increased as a result of the advertising campaign. If the level of significance is 0.05, what is the decision? 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 the mean is equal to $6,000 per day.
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4.A prior study(by Pew Research 2021)found that 35 out of 50 randomly selected US adults between the ages of 18 to 29 use (or have used) Instagram. Construct a 99% confidence interval for the population proportion of US adults between the ages of 18 to 29 use (or have used) Instagram. (round to three decimal places) a.Identify the following: X, N, P b.Which distribution should you use for this problem? c.What is the corresponding critical value for a 99% confidence interval? d.Calculate the margin of error (3 decimal places) e.Construct a 99% confidence interval f. Interpret the interval.
A. Find the sample meanB. Find the sample standard deviation C. Construct a 98% confidence interval for the population mean(please show what numbers you are putting into the formula for part b)
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?
Decide if the given claim represents the null or alternative hypothesis. a. A governor claimed that more than 76% of residents welcomed the stat-at-home order. The claim would be the alternative hypothesis. b. Hospitals claimed that they did not have the PPE required to keep their personnel safe. This could not be used as a null or alternative hypothesis. c. A Capitol High School spokesperson said that the SAT standard deviation was 22.5 points. The claim would be the alternative hypothesis. d. The senator claimed that unemployment in his state was greater than 18%. The claim would be the alternative hypothesis. e. She thought that on average she spent fewer than 12 hours a week on homework. The claim would be the null hypothesis.
A. You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 40 business days, the mean closing price of a certain stock was $107.60.Assume the population standard deviation is $10.03. The 90% confidence interval is B. A study on the personality characteristics of drug dealers sampled 100 convicted drug dealers and scored them each on the Wanting Recognition (WR) Scale, which provides a quantitative measure of a person's need for approval and sensitivity to social situations. (Higher scores indicate a greater need for approval.) The sample of drug dealers had a mean WR score of 44, with a standard deviation of 6. Use this information to find an interval estimate of the mean WR score for all convicted drug dealers. Use a confidence level of 99%. Interpret the result.…

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