An anger-management course claims that, after completing its seminar, participants will lose their tempers less often. Always a skeptic, you decide to test this claim. A random sample of 12 seminar participants is chosen, and these participants are asked to record the number of times that they lost their tempers in the two weeks prior to the course. After the course is over, the same participants are asked to record the number of times that they lost their tempers in the next two weeks. The following table lists the results of the survey. Using these data, test the claim at the 0.01 level of significance assuming that the population distribution of the paired differences is approximately normal. Let participants before completing the anger-management course be Population 1 and let participants after completing the anger-management course be Population 2.| Number of Times Temper Was Lost during a Two-Week Period ||----------------|---|---|---|---|---|---|---|---|---|---|---|---|---|| **Before** | 10| 6 | 5 | 8 | 8 | 7 | 7 | 10| 6 | 3 | 8 | 7 || **After** | 6 | 7 | 4 | 5 | 8 | 6 | 4 | 9 | 3 | 9 | 3 | 7 |**Step 3 of 3**: Draw a conclusion and interpret the decision.

In order to test the claim of anger-management course that the participants after completing the seminar on course will lose temper less often, a paired t-test needs to be conducted. The hypotheses test will contain two hypotheses: null hypothesis and alternate hypothesis. The alternate hypothesis is about the claim which is to be tested. If p-value<α, where, α is significance level, then null hypothesis can be rejected and it can be concluded that the test is significant. Assume, xd be the random variable representing the difference between number of times temper was lost during a 2-week period before and after the seminar. μd be the population mean difference between number of times temper was lost during a 2-week period before and after the seminar. x¯d be the sample mean difference between number of times temper was lost during a 2-week period before and after the seminar. sd be the sample standard deviation of differences. n be the number of pairs considered in sample. The null hypothesis and the alternate hypothesis for the test are, H0: μd=0 and Ha: μd>0, respectively.Prepare a table to show the differences. Number of times temper was lost Before the seminar x1 Number of times temper was lost After the seminar x2 Differences xd=x1-x2 10 6 4 6 7 -1 5 4 1 8 5 3 7 4 3 7 7 0 10 8 2 6 6 0 3 4 -1 10 9 1 3 3 0 8 7 1 1.083333 1.621354Calculate the value of x¯d by using the formula x¯d=∑i=1nxdin. Calculation gives x¯d≈1.083. Calculate the value of sd by using the formula sd=∑i=1nxdi-x¯d2n-1. Calculation gives sd≈1.621.The test statistic can be calculated by using the formula, t=x¯d-μd0sdn, where, μd0 denotes the mean difference as per null hypothesis. Substitute x¯d≈1.083, sd≈1.621, n=12 and μd0=0 in t=x¯d-μd0sdn and compute the value of t. t=1.083-01.62112≈2.314To find the p-value, use the t-distribution with degrees of freedom, d.f.=n-1 and evaluate the proportion of area that lies to the right of t=2.314 under the t-distribution curve with d.f.=11. This gives, p-value=0.0205. The significance level is 1%, that is, α=0.01. Since, p-value>α, the null hypothesis is not rejected. It can be said that there are insufficient evidence to conclude that the participants of seminar on course will lose temper less often, than earlier.