Answer to Question #279904 in Statistics and Probability for Nicole

Question

The manager of a restaurant in a large city claims that waiters working in all restaurants in his city earn an average of $150 or more in tips per week. A random sample of 25 waiters selected from restaurants of this city yielded a mean of $139 in tips per week with a standard deviation of $28. Assume that the weekly tips for all waiters in this city have a normal distribution.

 

a. Using the 1% significance level, can you conclude that the manager’s claim is true? Use the p value approach (20 Mks)

 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq 150$

$H_1:\mu<150$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=25-1=24$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -2.492159.$

The rejection region for this left-tailed test is $R = \{t: t < -2.492159\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{139-150}{28/\sqrt{25}}\approx-1.964286$$

Using the P-value approach:

The p-value for left-tailed, $df=24$ degrees of freedom, $t=-1.964286$ is $p = 0.03059,$ and since $p= 0.03059>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is less than $150,$ at the $\alpha = 0.01$ significance level.

Therefore, there is enough evidence to claim that the manager’s claim is true, at the $\alpha = 0.01$ significance level.