Answer to Question #349230 in Statistics and Probability for luck

Question

The cashier of a fastfood restaurant claims that the average amount spent by customers for dinner is P 120. 00. A sample

of 50 customers over a month period was randomly selected and it was found out that the average amount spent for dinner

was P 112. 50. Using a 0.05 level of significance, can it be concluded that the average amount spent by customers is more

than P 120. 00? Assume that the population standard deviation is P 6.50. (Use Critical Value Method)

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le120$

$H_1:\mu>120$

This corresponds to a right-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 right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z:z>1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{112.50-120}{6.50/\sqrt{50}}=-8.159$$

Since it is observed that $z=-8.159<1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z>-8.159)= 1,$ and since $p= 1>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is greater than 120, at the $\alpha = 0.05$ significance level.