Answer to Question #340704 in Statistics and Probability for Rose

Question

A mathematics teacher in senior high school developed a problem-solving test to randomly selected 40 grade 11 students. These students had an average score of 85 and a standard deviation of 5. If the population had a mean score of 90 and a standard deviation of 3, use 5% level of significance to test the hypothesis.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=90$

$H_1:\mu\not=90$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-90}{3/\sqrt{40}}\approx-10.541$$

Since it is observed that $|z| = 10.541 > 1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(z<-10.541)\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 90, at the $\alpha = 0.05$ significance level.