Answer to Question #210686 in Statistics and Probability for Cath

Question

For numbers 8-15: An association of City Mayors conducted a study to determine the average number of times a family went to buy necessities in a week. They found that the mean is 4 times in a week. A random sample of 20 families were asked and found a mean of 5 times in a week and a standard deviation of 2. Use 5% significance level to test that the population mean is not equal to 5. Assume that the population is normally distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=5$

$H_1:\mu\not=5$

This corresponds to a two-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 two-tailed test is $z_c=1.96.$

The rejection region for this two-tailed test is $R=\{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4-5}{2/\sqrt{20}}\approx-2.236$

Since it is observed that$|z|=2.236>1.96=z_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 5, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=2P(z<-2.236)=0.025347,$ and since $p=0.025347<0.05=\alpha,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 5, at the $\alpha=0.05$ significance level.