Answer to Question #342154 in Statistics and Probability for erika miguel

Question

Show your solution using the critical value method.

a previous study on the effect of blended learning module on senior high school student's achievement revealed that mean score pi is 85 with standard deviation of 3. juliana, a senior high school statistic teacher, surveyed a class of 30 students test the hypothesis at a 0.10 level of significance

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=85$

$H_a:\mu\not=85$

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 $\alpha = 0.10,$ $df=n-1=29$ degrees of freedom, and the critical value for a two-tailed test is $t_c =1.699127.$

The rejection region for this two-tailed test is $R = \{t: |t|>1.699127\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{83-85}{4\sqrt{30}}\approx-2.7386$$

Since it is observed that $|t|= 2.7386 >1.699127= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed $df=29$ degrees of freedom, $t=-2.7386$ is $p =0.00617,$ and since $p=0.00617<0.10=\alpha,$ 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 $\alpha = 0.10$ significance level.