Question

5. The owner of a comedy club has historically based her business decisions on the average age of her customers being 30 years. A random sample of customer ages is shown below. Based on this sample, is the average age greater than 30 years? Use a 5% level of significance. Assume that the distribution of ages is approximately normal. 37 30 26 35 45 42 51 40 43 27 39 38. 46 21 28 20 5. (a)

Answer 1

Here is the data:

Age 37 30 26 35 45 42 51 AC) 43 27 39 38 46 21 28 20 Count Average Standard Deviation 9.238 16 35.50

The provided sample mean is $\bar X = 35.50$ and the sample standard deviation is , and the sample size is .

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 30

Ha: μ > 30

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

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is tc​=1.753.

The rejection region for this right-tailed test is $R = {t: t \ge 1.753}$

(3) Test Statistics

The t-statistic is computed as follows:

$t = \frac{\bar X - \mu_0}{s/\sqrt{n}} = \frac{ 35.50 - 30}{ 9.2376/\sqrt{ 16}} = 2.382$

(4) Decision about the null hypothesis

Since it is observed that t=2.382>tc​=1.753, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0155, and since p=0.0155<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 30, at the 0.05 significance level.