Answer to Question #337333 in Statistics and Probability for Cyr

Question

The principal of tala senior highschool claims that the students in his school have above average intellegence. A random sample of 30 students IQ scores have a mean score of 113. The mean population IQ is 100 with a standard deviation of 15. Is there an evidence to support his claim?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le100$

$H_a:\mu>100$

This corresponds to a right-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 right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z > 1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{113-100}{15/\sqrt{30}}\approx4.7469$$

Since it is observed that $z = 4.7469 > 1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>4.7469)= 0.000001,$ and since $p= 0.000001<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 100, at the$\alpha = 0.05$ significance level.