Students in a statistics class at Penn State were asked, "About how many minutes do you typicallyexercise in a week?" Responses from the women in the class were as follows:60, 240, 0, 360, 450, 200, 100, 70, 240, 0, 60, 360, 180, 300, 0, 270Responses from the men in the class were as follows:180, 300, 60, 480, 0, 90, 300, 14, 600, 360, 120, 0, 2401. What assumptions or conditions are required if conclusions are to be made about amount ofexercise for the population of all Penn State students on the basis of these sample results?2. Assume that the conditions are met, and conduct a test to determine whether the mean amountof exercise differs for men and women. This is an independent samples t-test. Do not assumevariances are equal so use the unpooled formula. Specify the value of a that you chose to use.Report your t-test result in APA format: t(df) = p< .05 (or p > .05 if appropriate)

1) The assumptions necessary for performing the two-sample t-test, such that the results obtained from the sample data can be generalized for the entire target population are: The participants in the study should be randomly selected from the population. The two samples are taken from the normal distribution. The two samples are independent of each other. 2) The null hypothesis is a simple statement about the value of the population parameter which is liable to be rejected. It always contains an equal sign.An alternative hypothesis is simply the contradiction of the null hypothesis. It is the statement that is claimed by the researcher in the study.This is an example of a two-tailed test as the researcher is interested in knowing whether the mean amount of exercise between male and female students is different. State the null and the alternative hypothesis. H0:μw=μmH1:μw≠μmCompute the sample mean for the women group using the formula: x¯w=∑ixinw x¯w=60+240+0+,...,+27016 =289016 =180.625 Compute the sample mean for the men group using the formula: x¯m=∑ixinm x¯m=180+300+60+,...,+24013 =274413 =211.077Compute the sample variance for the women group using the formula: s2w=∑ixi-x¯wnw-1 s2w=60-180.6252+240-180.6252+0-180.6252+,...,+270-180.625216-1 =14550.39+3525.39+32625.39+,...,+7987.8915 =312293.7515 =20819.58 Compute the sample variance for the men group using the formula: s2m=∑ixi-x¯mnm-1 s2m=180-211.0772+300-211.0772+60-211.0772+,...,+240-211.077213-1 =965.78+7907.31+22824.24+,...,+836.5412 =437100.9212 =36425.08Compute the sample test statistic using the formula: t=x¯w-x¯ms2wnw+s2mnm t=180.625-211.07720819.5816+36425.0813 =-30.45264.056 =-0.475Compute the degrees of freedom using the formula: df=s2wnw+s2mnm2s2wnw2nw-1+s2mnm2nm-1 df=20819.5816+36425.0813220819.5816216-1+36425.0813213-1 =21.95 ≃22 By using the t-distribution table for the two-tailed test, the p-value for the 22 degrees of freedom and corresponding to the value of the test statistic t=-0.475 is, 0.6395. Since the p-value of the test 0.6395 is greater than the assumed level of significance 0.05; therefore, the null hypothesis is failed to reject. Hence, at the 5% level of significance, there is insufficient evidence to conclude that the mean amount of exercise differs for men and women.