Answer to Question #334186 in Statistics and Probability for KLTino

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 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 4. Assume that the population is normally distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=4$

$H_a:\mu\not=4$

This corresponds to a two-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 $\alpha = 0.05,$ $df=n-1=19$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.093024.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.093024\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{5-4}{2/\sqrt{20}}\approx2.236068$$

Since it is observed that $|t| = 2.236068 >2.093024=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=19$ degrees of freedom, $t=2.236068$ is $p=0.037541,$ and since $p=0.037541<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 4, at the $\alpha = 0.05$ significance level.