Question

Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the six months following the exercise program. Below are the results. Employee Before After 0 JOU WN nin mnm Click here for the Excel Data File At the 0.01 significance level, can he conclude that the number of absences has declined? Estimate the p-value. a. State the decision rule for 0.010 significance level. (Round your answer to 3 decimal places.) Reject Ho ift > b. Compute the test statistic. (Round your answer to 3 decimal places.) The test statistic is c. State your decision about the null hypothesis. Do not reject Ho Reject Ho

Answer 1

Given the sample, Befose : 657 6 4 6 5518.5 627 5 3 2 372375 (%) Roxi-Yi : 4- 5-1 3.32 lá. we obtained sample means, $-5.5, -3.75 sample standard deviations, Sx=0.926 S, = 1.982 I see J 1 1 8(xi-2² I sample size, nes we obtained, means of differences, ac 1.55 Standard deviation of differences, Se - 2.435 since we have to test whether no. of absentees has declined, we will test, whether anerage no of. absenters before exercise program is greater than offer the hathat of after the aluse program. Therefore one null hypothesis, Ho: foto (means differences are equal) against the allee native, Hi Horo (Mx > Hp) This corresponds to a oright failed test for which t test for paired samples - Rejection region and decision onile. . Based on the information provided, for significance level, d=o.ol, and degrees of freedom odf=7.(n-128-1) the critical value for right taled bt. te=29998 (too, y = 2.995, from t table) is used

. Therefore the rejection region is, - R={tity 2.9982 <b> test statistic, 1-75 2-435 - 2.033 since it is observed that the24033 b bz2.033 <te= 2.998, it is I concluded that the null hypothesis I is not rejected . Now pvalue, p= P(t <tcl (from t = 0.0408 poobabilitas i p value, p7,0.01. 1 = null heppothesis is not rejected (e) Do not reject Ho. It is concluded that, the roof absentes diants destined there is no evidesse i to claim that the no of absentees is declined.