Answer to Question #342464 in Statistics and Probability for Ari

Question

A shop owner claims that on average, his shop has a daily turnover of R75 000. Test this claim at the 5% significance level, if the average daily turnover for a sample of 88 days was R71 500. Assume a population standard deviation of R4 450.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=75000$

$H_1:\mu\not=75000$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{71500-75000}{4450/\sqrt{88}}\approx-7.3782$$

Since it is observed that $|z|=7.3782>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-7.3782)=0,$ and since $p= 0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 75000, at the $\alpha = 0.05$ significance level.