please include the confidence interval and the P-value for each problem and which hypothesis to accept.and please use the calculator method not the book or excell ti 83 or 84 ### Student Ratings of Trustworthiness for Mitt Romney**Scenario**Twenty-nine college students with a positive outlook on Mitt Romney compared to Barack Obama in the 2012 presidential election were asked to evaluate Romney's trustworthiness based on their mental image of his face. Ratings were given on a scale from 0 to 7:- 0: Not at all trustworthy- 7: Extremely trustworthy#### RatingsHere are the 29 ratings provided by the students:2.6, 3.2, 3.7, 3.3, 3.4, 3.6, 3.7, 3.8, 3.9, 4.1, 4.2, 4.9, 5.7, 4.2, 3.9, 3.2, 4.5, 5.0, 5.0, 4.6, 4.6, 3.9, 3.9, 5.3, 2.8, 2.6, 3.0, 3.3, 3.7### Tasks(a) **Stem-and-Leaf Plot Analysis**Create a stemplot to visualize the data. Analyze the plot to assess any major departures from normality.(b) **Confidence Interval**Calculate a 95% confidence interval for the mean rating.(c) **Statistical Significance**Determine if the mean rating is significantly greater than 3.5 at the 5% significance level.> These tasks aim at assessing perceptions of trustworthiness and analyzing them statistically.

Name: please include the confidence interval and the P-value for each proble
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: please include the confidence interval and the P-value for each problem and which hypothesis to accept.and please use the calculator method not the book or excell ti 83 or 84 ### Student Ratings of Trustworthiness for Mitt Romney**Scenario**Twenty-ni

a) Step by step procedure to draw stem-and leaf plot using MINITAB software: Select Graph > Stem and leaf. Select the column of variables in Graph variables. Enter increment value as 0.5 Select OK. Output of above procedure is as follows: The stem and leaf plot shows that the data is normally distributed.b) Here, the sample size is 29 that is reasonably small. The population standard deviation is unknown here. Therefore, z-test cannot be used here. In this situation student’s t-distribution can be used. Let µ be the true mean rating. Let x bar be the sample mean rating. s be the sample standard deviation. n is the sample size. Here the confidence interval is 95%. Therefore, α = 0.05. The degrees of freedom, n - 1 = 29 - 1 =28, From the EXCEL, using the formula, =T.INV.2T(0.05,28), the critical value is 2.0484. From the EXCEL, the mean of the data is 3.9172 and the standard deviation is 0.796. Therefore the 95% confidence interval of mean rating is, The 95% confidence interval of mean rating is between 3.55 and 4.2844. c. Null hypothesis: H0 : μ = 3.5 That is, the mean rating is 3.5. Alternative hypothesis: H1 : μ > 3.5 That is, the mean rating is more than 3.5. Test statistic: The test statistic for t test is given by: Here, n = 29 and μ0 = 3.5. From Excel the mean and standard deviation for the data is 3.9172 and 0.796 respectively. The test statistic for hypothesis test for mean is given by: Decision rule: If t ≥ tα,n-1 ,reject the null hypothesis. Otherwise fail to reject the null hypothesis. From the EXCEL, using the formula, =T.INV(0.95,28), the critical value is 1.7011. Here the test statistic, 2.819 is greater than the critical value, 1.7011. Then, the null hypothesis will be rejected. Therefore, there is a significance evidence at 5% level of significance that the mean rating is greater than 3.5.