The average student loan debt is reported to be $25,235. A student belives that the student loan debt ishigher in her area. She takes a random sample of 100 college students in her area and determines themean to be $27,524 and the standard devition to be $6000. Is there sufficient evidence to support thestudent' claim at a 5% significance level?a) Determine the null and alternative hypotheses.Ho: H =Ha: µ Select an answer(Put in the correct symbol and value)b) Determine the test statistic. Round to two decimals.t =c) Find the p-value. Round to 4 decimals.P-value =d) Make a decision.O Reject the null hypothesisO Fail to reject the null hypothesise) Write the conclusion.O There is not sufficient evidence to support the claim that student loan debt is higher than $25,235.O There is sufficient evidence to support the claim that student loan debt is higher than $25,235.

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Answered: The average student loan debt is…

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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The Gallup Poll asked 1015 U.S. adults whether they believed that people should pay sales tax on items purchased over the internet. Of these, 477 said they supported such a tax. Does the survey provide convincing evidence that less than 42% of U.S. adults favor an internet sales tax? Use the 0.05 level of significance. 1. STATE THE NULL AND ALTERNATIVE HYPOTHESIS: 2. T STATISTIC/ P VALUE 3.REJECT OR FAIL THE NULL HYPOTHESIS 4. IS THE CLAIM SUPPORTED AT THE 0.05 LEVEL OF SIGNIFICANCE?
The Survey of Study Habits and Attitudes (SSHA) is a psychological test with scores that range from 0 to 200. The mean score for U.S. college students is 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 students from the more than 1000 students at her college who are at least 30 years of age. The teacher wants to perform a test at the a = 0.05 significance level of Ho μ = 115 Ha: μ> 115 where μ = the mean SSHA score in the population of students at her college who are at least 30 years old. The sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. Find and interpret the P-value. The P-value is 0.0202. Assuming that the true mean SSHA score is 115, there is a 0.0202 probability of getting a sample mean SSHA score at least as far as 125.7 (in either direction) just by chance in a random sample of 45 students. The P-value is 0.0101. Assuming that the true mean SSHA score is 115, there is a 0.0101…
Part: 2/5 Part 3 of 5 (c) Compute the P-value. Round the answer to at least four decimal places. P-value= X Ś
Test the claim that the proportion of men who own cats is significantly different than 20% at the 0.02 significance level. The null and alternative hypothesis would be: Ho:µ = 0.2 Ho:H = 0.2 Ho:p= 0.2 Ho:µ = 0.2 Ho:p= 0.2 Ho:p= 0.2 H1: µ > 0.2 H1:470.2 H1:p7 0.2 H1:µ 0.2 The test is: left-tailed right-tailed two-tailed Based on a sample of 4 people, 25% owned cats The test statistic is: (to 2 decimals) The positive critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Check Answer
Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below. H0:μ=8.2 seconds; Ha:μ<8.2 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.
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Use the departure delay times and the arrival delay times to test the claim that there is a correlation between departure delay times and arrival delay times. Test for rank correlation with a 0.05 significance level. Click here to view the flight delay data. Click here to view the critical values of Spearman's rank correlation coefficient. Flight delay data .... Determine the null and alternative hypotheses for this test. Choose the correct answer below. O A. Hg: Ps =0 Departure Delay Arrival Departure Delay Arrival O B. Ho: Ps #0 H: Ps = 0 Departure Delay Arrival H: Ps #0 Delay Delay Delay - 36 -7 11 14 OC. Ho:r 0 H:r=0 O D. Ho: r =0 H:rg 40 -21 15 25 -33 10 -32 Determine the the correlation coefficient. - 5 32 13 4 12 16 -23 (Round to three decimal places as needed.) 3 14 -38 60 28 13 3 13 149 111 Determine the critical value(s) of the correlation coefficient, -38 -7 -35 - 3 -21 -23 23 - 2 -10 14 (Round to three decimal places as needed. Use a comma to separate answers as needed.)…
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