Answer to Question #214291 in Statistics and Probability for Winnie kea Lobas

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:

Null hypothesis: the population mean is equal to 90.

Alternative hypothesis: the population mean is not equal to 90.

$H_0:\mu=90$

$H_1:\mu\not=90$

The significance level is $\alpha=0.05.$

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

The critical value for a two-tailed test is $z_c=z_{1-\alpha/2}=1.96.$

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

The z-statistic is computed as follows:

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

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

Using the P-value approach: The p-value is $p=2P(Z<-10.541)=0,$ and since $p=0<0.05=\alpha,$ 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.