Answer to Question #207959 in Statistics and Probability for Lovely

Question

An association of city mayons 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 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:

$\mu=5$

$\mu\not=5$

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 $\alpha=0.05$ and $df=n-1=40-1=39$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.022691.$

The rejection region for this two-tailed test is $R=\{t: |t|>2.022691\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4-5}{2/\sqrt{40}}=-3.162278$$

Since it is observed that $|t|=3.162275>2.022691=t_c,$ 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 for two-tailed, $\alpha=0.05, df=39,$

$t=-3.162278,$ is $p=0.003027,$ and since $p=0.003027<0.05=\alpha,$ 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.