Question

(a) state the null and alternative hypotheses, (b) compute the test

statistic, (c) determine the critical value and sketch the rejection region and non-rejection

region in the normal curve.

1. The cahier of a fast-food restaurant claims that the average amount spent by customers for dinner is

Php 120.00. A sample of 50 customers over a month was randomly selected and it was found out

that the average amount spent for dinner was Php 122.50. Construct the critical regions using a 0.05

level of significance to conclude that the average amount spent by customers is more than Php

120.00. Assume that the population standard deviation is Php6.50.

Answer 1

1.

a) The following null and alternative hypotheses need to be tested:

$H_0:\mu=120$

$H_1:\mu>120$

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

b) The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{122.50-120}{6.50/\sqrt{50}}\approx2.7196$$

c) 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 $R = \{z:z> 1.6449\}.$

d) Since it is observed that $z=2.7196>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>2.7196)=0.003268,$ and since $p=0.003268<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

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