Answer to Question #350771 in Statistics and Probability for john

Question

A canteen owner claims that the average meal cost of his usual clients is

Php180. In order to test his own claim, he took a random sample of 30

receipts and computed the mean cost of Php210 with a standard deviation of

Php25. Test the hypothesis at 0.01 level of significance. Show all steps

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=180$

$H_1:\mu\not=180$

This corresponds to a two-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 $\alpha = 0.01,$ $df=n-1=29$ and the critical value for a two-tailed test is $t_c = 2.756386.$

The rejection region for this two-tailed test is $R = \{t:|t|> 2.756386\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{210-180}{25/\sqrt{30}}=6.57267$$

Since it is observed that $|t|=6.57267> 2.756386=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=29$ degrees of freedom, $t=6.57267$ is $p=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 180, at the $\alpha = 0.01$ significance level.