Answer to Question #305006 in Statistics and Probability for Alissa

Question

Last year the employees of the city health department donated an average of $ 8 to the rescue squad. Test the hypothesis at the 0.01 level of significance that the average contribution this year is still $ 8 if a random sample of 35 employees showed an average donation of $ 8.90 with a standard deviation of $ 1.75.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=8$

$H_1:\mu\not=8$

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 $\alpha = 0.01,$ $df=n-1=34$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.728394.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.728394\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{8.90-8}{1.75/\sqrt{35}}\approx3.042555$$

Since it is observed that $|t| = 3.042555 > 2.728394=t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=34$ degrees of freedom, $t=3.042555,$ is $p=0.0045,$ and since $p=0.0045<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 8, at the $\alpha = 0.01$ significance level.