A personnel manager reads in a national newspaper that the average office worker

Question

A personnel manager reads in a national newspaper that the average office worker wastes 4.5 hours per week by arriving late, socializing, conduct personal business, working slow, faking illness, taking long lunch hours, and sleeping on the job. By observing his office staff, he obtains a random sample for a hypothesis test because he believes that his staff is different. The average time wasted per week for 10 weeks was 4.1 hours with a standard deviation of 1.33 hours. If a hypothesis test was performed at the 0.01 level with an assumption of a normally distributed population, the test statistic and p-values are:

Explanation / Answer

Null and Alternative Hypothesis:

H0: µ = 4.5

H1: µ ¹ 4.5

Level of significance = .01

Test Statistics:

t = ( x bar – Mew)/(s/sqrt(n))

    = (4.1 -4.5)/(1.33/sqrt(10))

   = -0.951

Now we use excel function to find the p-value.

Here the degrees of freedom = n – 1 = 10 – 1 = 9

=TDIST(0.951,9,2)

= 0.3664

The test statistics is -0.951 and the p-value is 0.3664

Here the p-value is greater than the level of significance (0.3664>.01). We fail to reject the null hypothesis.

At 1% level of significance there is not sufficient evidence to conclude that his staffs are different in respect of weekly time wasted compare to national average value.

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