Answer to Question #345312 in Statistics and Probability for Rhea

Question

Activity 3 region

Find the critical value of each given problem. Skatch the curve and shade the wjection

1 14-90

The sample mean is 69 and simple size is 35 The population follows a normal distritsation with standard deviation 5 Usca w 0.05

2. A redaurant cashier claimed that the average amount spent by the endomers for dinner is P125.00 Over a month period, a sample of 50 customers was selected and it was found that the everage amount spent for dinner was P130.00 Using 005 level of significance can it be concluded that the average amount spent by customers is more than P125.00 Astose that the population standard deviation is 7.00

Answer 1

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

$$|z|\ge1.96$$

2. The following null and alternative hypotheses need to be tested:

$H_0:\mu\le125.00$

$H_1:\mu>125.00$

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 left-tailed test is $R = \{z:z>1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{130-125}{7/\sqrt{50}}\approx5.05$$

Since it is observed that $z=5.05>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>5.05)=0,$ 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 greater than 125, at the $\alpha = 0.05$ significance level.