Answer to Question #241076 in Statistics and Probability for Aven

Question

It is known from the records of the city schools that the standard deviation of mathematics test scores on the XYZ test is 5. A sample of 200 pupils from the system was taken and it was found out that the sample mean score is 75. Previous tests showed the population mean to be 70. Is it safe to conclude that the sample is significantly different from the population?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu=70$

$H_1:\mu\not=70$

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:

$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{75-70}{5/\sqrt{200}}=10\sqrt{2}$

$\approx14.142$

Since it is observed that $|z| = 14.142 >1.96= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed $\alpha=0.05,$ $z=14.142$ is $p=2P(z>14.142)\approx0,$ 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 sample is significantly different from the population, at the $α=0.05$ significance level.