Hollar is vice president for human resources for a large manufacturing company.

Question

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 paticipants 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. Click here for the Excel Data File At the 010 significance level, can he conclude that the number of absences has declined? Estimate the pvalue. a. State the decision rule for 0100 significance level. (Round your answer to 3 decimal places.) eject HO ft 1.415 Prex 30f 8..ill.. Next >

Explanation / Answer

Given that,

null, H0: Ud > 0
alternate, H1: Ud < 0
level of significance, = 0.1
from standard normal table,left tailed t /2 =1.415
since our test is left-tailed
reject Ho, if to < -1.415
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 1.88
We have d = 1.88
pooled variance = calculate value of Sd= S^2 = sqrt [ 57-(15^2/8 ] / 7 = 2.03
to = d/ (S/n) = 2.62
critical Value
the value of |t | with n-1 = 7 d.f is 1.415
we got |t o| = 2.62 & |t | =1.415
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < 2.6181 ) = 0.98275
hence value of p0.1 < 0.98275,here we reject Ho
ANSWERS
---------------
null, H0: Ud > 0
alternate, H1: Ud < 0
test statistic: 2.62
critical value: reject Ho, if to < -1.415
decision: Reject Ho
p-value: 0.98275

X Y X-Y (X-Y)^2 6 2 4 16 7 6 1 1 7 4 3 9 6 7 -1 1 4 5 -1 1 7 3 4 16 5 3 2 4 5 2 3 9 15 57

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