Answer to Question #342114 in Statistics and Probability for erika miguel

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.