ReportOneway Analysis of Movies in Last Year By GenderOneway Analysis of Movies in Last Year By GenderPooled t Test25Summary of FitRsquare0.053085Adj RsquareRoot Mean Square ErrorMean of ResponseObservations (or Sum Wgts)0.0281884.6581859.0254010-t TestM-FAssuming equal variancesMDifference2.1500 t Ratio1.459559GenderStd Err Dif1.4730DF38Upper CL Dif5.1320 Prob > |E|0.1526Lower CL Dif -0.8320 Prob > t0.0763Pooled t TestConfidence0.95 Prob |t|Difference1.459559Std Err Dif38Std Error uses a pooled estimate of error varianceUpper CL Dif0.1526Lower CL Dif -0.3335 Prob >t0.0763t TestConfidence0.9 Prob |E|Difference1.459550Analysis of VarianceStd Err Dif37.80348Upper CL Dif0.1527SourceDF Sum of Squares Mean Square FRatio Prob >FLower CL Dif -0.8325 Prob > t0.0763Gender146.2250046.2250 2.13030.1526Confidence0.95 Prob |t|Lower CL Dif -0.3338 Prob >t0.9 Prob O (a)A movie theater is interested in seeing if female or males, on average, attend the movies a different amount. What is the hypothesis test to be done? (Use u, - Hg, where u, ismales and u, is females.)O Ho: H1 = H2O Ho: H, = 0.25H1 42 = 0.75O Ho: H1 = H2O Ho: H1> H2O (b)Assume the population variances for the two samples are equal (then df = n1 + n2 - 2). Find the t-statistic, critical value, and the p-value associated with the hypothesis test atthe 90% confidence level. (Round your t-statistic and p-value to four decimal places. Round your critical value to three decimal places.) (Use u, - lg, where u, is males and u, isfemales.)t-statisticcritical valuep-valueO (c)At a 90% confidence level, should the null hypothesis be accepted or rejected? Fill in the blanks.The null hypothesis should ---Select--. The number of times, on average, that females and males attend the movies each year ---Select-- v statistically different.(d)Assume the population (all people) standard deviation is known to be four movies per year each for males and females. What are the new test statistic and critical value for a 90%confidence level? (Round your answers to three decimal places.)test statisticcritical valueYou may need to use the appropriate table in the Appendix of Tables to answer this question.

Note: Hey there! Thank you for the question. As you have posted multiple sub-parts, according to our policy, we have solved the first three for you. If you need help with the other parts, please re-post the question with complete background information, and mention the parts.a. Denote μ1, μ2 as the true population mean numbers of times that males and females attended the movies in the previous years, respectively. As it is of interest to know whether the mean is different for males or females, the null and alternative hypotheses are as follows: H0: μ1 = μ2 versus H1: μ1 ≠ μ2.b. For the t-test at 90% confidence level assuming equal or pooled variances, consider the following section from output given in your question: For a 100 (1 – α) % confidence interval, the confidence level is 100 (1 – α) % or (1 – α). For a 90% confidence level, 100 (1 – α) = 90, so that (1 – α) = 0.9. Note that the confidence level, given as "Confidence" in the output is 0.9. Further, the top of the table reveals that it is a t-test for "M-F", that is, males minus females, and it assumes equal variances. As a result, this section of the output is suitable for the question. The t-statistic is given as "t Ratio". The value of the t-statistic (rounded to four decimal places) is 1.4596. The alternative hypothesis reveals that this is a two-tailed test. As a result, the p-value to be used is the one given as "Prob > |t|". The value of the p-value (rounded to four decimal places) is 0.1526. The degrees of freedom, given as "DF", is 38. The critical value can be obtained using the Excel formula: =T.INV.2T(0.1,38), which gives the value 1.686. Note that 0.1 in the formula is the value of α; for the 90% confidence level, since (1 – α) = 0.9, α = 0.1. This formula shows that 1.686 is the positive value of the t-variable with 38 degrees of freedom, above which, α/2 = 0.05 proportion of the observations lie; below -1.686, the remaining α/2 = 0.05 proportion of the observations lie. Hence, the critical value (rounded to three decimal places) is 1.686.c. Decision rule using p-value: Reject H0 at significance level α, if p-value ≤ α. Otherwise, fail to reject H0. Since p-value (= 0.1526) > α (= 0.10), fail to reject H0. Thus, the null hypothesis should not be rejected. [If the options in your question do not include "fail to reject" but use the word accept, then select that option. But by convention, we do not "accept" the null hypothesis, we merely fail to reject it, when the p-value is greater than the significance level] The number of times, on an average, that females and males attend the movies each year are not statistically different.