Answer to Question #350597 in Statistics and Probability for secret

Question

Conduct a t-test for the null hypothesis against

alternative hypothesis based on 16 random

observations. The sample mean is 110 and the sample standard

deviation is 10. Use α = 0.05.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=100$

$H_1:\mu>100$

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 $\alpha = 0.05,$ $df=n-1=15$ and the critical value for a right-tailed test is $t_c =1.75305.$

The rejection region for this right-tailed test is $R = \{t:t>1.75305\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{110-100}{10/\sqrt{16}}=4$$

Since it is observed that $t=4>1.75305=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=15$ degrees of freedom, $t=4$ is $p=0.00058,$ and since $p=0.00058<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 100, at the $\alpha = 0.05$ significance level.