Answer to Question #344869 in Statistics and Probability for Angie

Question

Hi:u=85 Ha:u=85

The sample mean is 83,the sample size is 39,and the standard deviation is 5. Use a=0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=85$

$H_1:\mu\not=85$

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 $\alpha = 0.05,$ $df=n-1=38$ and the critical value for a two-tailed test is $t_c =2.024394.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.024394\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{83-85}{5/\sqrt{39}}=-2.498$$

Since it is observed that $|t|=2.498>2.024394=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=38$ degrees of freedom, $t=-2.498$ is $p= 0.016935,$ and since $p= 0.016935<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 85, at the $\alpha = 0.05$ significance level.