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 1 6 5 2 6 2 3 7 1 4 7 3 5 4 3 6 3 6 7 5 3 8 6 7 At the 0.05 significance level, can he conclude that the number of absences has declined? Estimate the p-value. A. State the decision rule for 0.05 significance level. (Round your answer to 3 decimal places.) B. Compute the test statistic. (Round your answer to 3 decimal places.) C. The p-value is D. State your decision about the null hypothesis.

          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 1 6 5 2 6 2 3 7 1 4 7 3 5 4 3 6 3 6
7 5 3 8 6 7 At the 0.05 significance level, can he conclude that
the number of absences has declined? Estimate the p-value. A. State
the decision rule for 0.05 significance level. (Round your answer
to 3 decimal places.) B. Compute the test statistic. (Round your
answer to 3 decimal places.) C. The p-value is D. State your
decision about the null hypothesis.

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Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Adi S

04/30/2022

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The alternative hypothesis (Ha) is that the mean number of absences after the fitness program is less than the mean number of absences before the fitness program. Show more…

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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 1 6 5 2 6 2 3 7 1 4 7 3 5 4 3 6 3 6 7 5 3 8 6 7 At the 0.05 significance level, can he conclude that the number of absences has declined? Estimate the p-value. A. State the decision rule for 0.05 significance level. (Round your answer to 3 decimal places.) B. Compute the test statistic. (Round your answer to 3 decimal places.) C. The p-value is D. State your decision about the null hypothesis.

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00:01 It is given that to examine the fitness program as the attendant rate is declined or not for n -equal to 8 employees there before and after attendance of before and after attending the fitness program given are 6, 6, 6, 7, 7, 7, 4, 3, 5, 6 and the after attendance are respectively 5, 2, 1, 3, 3, 6, 3, 7.

00:44 Let us take the difference, d .i equal to before, attendance, b, minus after attendance a.

00:53 And the values are 1 4 6 4 1 minus 3 2 and minus 1 now sum of all these values is 14 now the mean of the difference d bar equal to summation di by n which is 14 by 8 simplifying this gives d -bar equal to 1 .75.

01:30 Now, the value of d -i minus d -bar the old square is 0 .56 to 5 .0 .6 to 5 .8 .06 to 5 .8 .06 to 5 .5.

01:53 0 .0625, 0 .5625.

02:00 22 .5625, and 0 .0625 and 7 .5625.

02:13 Sum of all this values is 59 .5.

02:18 Now the standard deviation of the distance sd is calculated as summation d i minus d bar the old square divided by n minus 1 which is 59 .5 divided by 8 minus 1 simplifying this gives 2 .91 double 5 as the part a the decision rule is given as below if the calculated p value is less than given significance level of 0 .05 then the null hypothesis will be rejected and if the calculated p value is greater than 0 .05 then the null hypothesis is not rejected the null hypothesis h0 for this problem is mu -d, that is, mean of the difference, equal to 0, and the alternative hypothesis, h -a is such that mu -d is greater than 0.

03:29 Thus, the decision rule is provided as below.

03:36 In part b, the test statistics, t is calculated as d -bar, divided by sd by root 10.

03:48 Substituting the values gives 1 .75 divided by 2 .991 double 5 divided by root of 8...

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