Answer to Question #276554 in Statistics and Probability for Shen

Question

John and his friend were recently having a heated discussion about the number of men living in a western parish who possibly cared for children that  are not biologically theirs during 2020. Assume that a survey of the parish’s population revealed that the average number of cases and the standard deviation was 5,000 and 6000 in 2019, respectively.

John and his friend do not believe that the population average is correct. Assume that a survey based on a sample of the 800 people was conducted and found that the average number of cases in the western parish was 10,000 over the 12 months period. Conduct a hypothesis test, using a 5% significance level to determine if the average number of cases is actually higher based on the sample data.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq5000$

$H_1:\mu>5000$

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:

$$z=\dfrac{x-\mu}{\sigma/\sqrt{n}}=\dfrac{10000-5000}{6000/\sqrt{800}}\approx23.5702$$

Since it is observed that $z = 23.5702 >1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>23.5702)\approx0,$

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 5000, at the $\alpha = 0.05$ significance level.