Answer to Question #347947 in Statistics and Probability for Criceljoice

Question

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 25 families were asked and found a mean of 5 times in a week and a standard deviation of 2. Use 1% significance level to test that the population mean is greater than 4 hours. Assume that the population is normally distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le4$

$H_1:\mu>4$

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 $\alpha = 0.01,$ $df=n-1=24$ and the critical value for a right-tailed test is $t_c =2.492159.$

The rejection region for this right-tailed test is $R = \{t:t>2.492159\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{5-4}{2/\sqrt{25}}=2.5$

Since it is observed that $t=2.5>2.492159=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=24$ degrees of freedom, $t=2.5$ is $p=0.009827,$ and since $p= 0.009827<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 4, at the $\alpha = 0.01$ significance level.