**Do college students enjoy playing sports just as much as watching sports?** A researcher randomly selected ten college students and asked them to rate playing sports and watching sports on a scale from 1 to 10, with 1 meaning they have no interest and 10 meaning they absolutely love it. The results of the study are given below.**Playing vs. Watching Sports Data**| Play | Watch ||------|-------|| 4 | 1 || 2 | 1 || 1 | 1 || 2 | 1 || 7 | 5 || 8 | 4 || 1 | 3 || 1 | 3 || 3 | 1 || 6 | 1 |Assume a Normal distribution. What can be concluded at the α = 0.05 level of significance?For this study, we should use [Select an answer]a. The null and alternative hypotheses would be:- \(H_0\) : [Select an answer] [Select an answer] [Select an answer] (please enter a decimal)- \(H_1\) : [Select an answer] [Select an answer] [Select an answer] (Please enter a decimal)b. The test statistic \( \text{?} \) = [ ] (please show your answer to 3 decimal places.)c. The p-value = [ ] (Please show your answer to 4 decimal places.)d. The p-value is \( \text{?} \) αe. Based on this, we should [Select an answer] the null hypothesis.f. Thus, the final conclusion is that...- [ ] The results are statistically insignificant at α = 0.05, so there is statistically significant evidence to conclude that the population mean rating for playing sports is equal to the population mean rating for watching sports. - [ ] The results are statistically significant at α = 0.05, so there is sufficient evidence to conclude that the ten students that were surveyed rated playing sports the same as watching sports on average.- [ ] The results are statistically significant at α = 0.05, so there is sufficient evidence to conclude that the population mean rating for playing sports is not the same as the population mean rating for watching sports.- [ ] The

From the provided information, Sample size (n) = 10 Level of significance (α) = 0.05 The difference table is as follow: 3 1 0 1 2 4 -2 -2 2 5For this study we should use paired t test. a) The null and alternative hypotheses are as follow: H0: µd = 0 H1: µd ≠ 0 The mean and standard deviation of difference can be obtained as: xd=∑xdn=3+⋯+510=1.4s=∑xd-xd2n-1=3-1.42+⋯+5-1.4210-1=2.319b) The test statistic can be obtained as: t=xd-μdsdn=1.4-02.31910=1.909 The test statistic is 1.909.c) Degrees of freedom = n – 1 = 10 - 1 = 9 The p value of test statistic at 9 degrees of freedom can be obtained using excel function as: =T.DIST.2T(1.909,9) The excel output is as follow: Thus, the required p value is 0.0886 (rounding four decimal).d) The p value is greater than level of significance (α). e) Based on this, we should fail to reject the null hypothesis. f) Thus, the final conclusion is that the results are statistically insignificant at α = 0.05, so there is insufficient evidence to conclude that the population mean rating for playing sports is not the same as the population mean rating for watching sports.