Answer to Question #297549 in Statistics and Probability for jom

Question

It is known from the records of the city schools that the standard deviation of mathematics test score on the XYZ test is 5. A sample of 200 pupils were taken and it was found 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 at 5% significance level?

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 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 population mean $\mu$  is different than 70, at the $\alpha = 0.05$ significance level.