Answer to Question #347846 in Statistics and Probability for Justine Jose

Question

A researcher used a developed problem solving test to randomly select 50 Grade 6 pupils. In

this sample, and . The mean and the standard deviation of the population used in

the standardization of the test were 75 and 15, respectively. Use the 95% confidence level

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=75$

$H_1:\mu\not=75$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{80-75}{15/\sqrt{6}}=0.8165$$

Since it is observed that $z=0.8165<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(z>0.8165)= 0.414214,$ and since $p=0.414214>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 75, at the $\alpha = 0.05$ significance level.