Answer to Question #140424 in Statistics and Probability for Esther Gathenya

Question

The labour charge for repairs at an automobile service center are
based on a standard time specified for each type of repair. The time
specified for replacement of universal joint in a drive shaft is one
hour. The manager reviews a sample of 30 such repairs. The average
of the actual repair times is 0.86 hour with standard deviation 0.32
hour.
a. Test at the 1% level of significance the null hypothesis that the
actual mean time for this repair differs from one hour.
b. The sample mean is less than one hour, suggesting that the mean
actual time for this repair is less than one hour. Perform this test, also
at the 1% level of significance. (The computation of the test statistic
done in part (a) still applies here.)

Answer 1

1. $H_0:\mu=1$

$H_1:\mu\ne1$

$Tc=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}$

$=\frac{0.86-1}{\frac{0.32}{\sqrt{30}}}=-2.4258$

$Cv=t_{{\frac{\alpha}{2}},n-1}$

$=t_{0.005,29}=2.756$

Since the absolute test statistic tc=2.4258 is less than the critical value cv =2.756, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that the actual mean differs from one.

2. $H_0:\mu=1$

$H_1:\mu<1$

$Tc=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}$

$=\frac{0.86-1}{\frac{0.32}{\sqrt{30}}}=-2.4258$

$Cv=t_{{\alpha},n-1}$

$=t_{0.01,29}=2.462$

Since the absolute test statistic tc=2.4258 is less than the critical value cv =2.462, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that the actual mean is less than one.

The computation of the test statistic in part (a) still applies in part (b) since the change in the alternative hypothesis only affects the critical value.