Answer to Question #346277 in Statistics and Probability for hhhh

Question

 Identify the critical value of each given problem. Find the rejection region and sketch the curve on a separate sheet of paper.

“According to the radio announcer, the average price of kilogram of pork liempo is more than ₱210.00. However, a sample of 15 prices randomly collected from different markets showed an average of ₱215.00 and standard deviation of ₱9.00. Using 0.05 level of significance, is there sufficient evidence to conclude that the average price of pork liempo is more than ₱210.00?”

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le210$

$H_1:\mu>210$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=14$ and the critical value for a right-tailed test is $t_c = 1.76131.$

The rejection region for this right-tailed test is $R = \{t:t> 1.76131\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{215-210}{9/\sqrt{15}}=2.1517$$

Since it is observed that $t=2.1517> 1.76131=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=14$ degrees of freedom, $t=2.1517$ is $p= 0.02468,$ and since $p= 0.02468<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 210, at the $\alpha = 0.05$ significance level.