Answer to Question #348114 in Statistics and Probability for Trisha

Question

The principal claims that the students in his school are above average intelligence. A random sample of 35 IQ scores have a mean score of 110.5. Is there sufficient evidence to support the principal’s claim? The mean population IQ is 100 with a standard deviation of 25. Use =1%.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le100$

$H_1:\mu>100$

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 $\alpha = 0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263.$

The rejection region for this right-tailed test is $R = \{z:z>2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{110.5-100}{25/\sqrt{35}}=2.4848$$

Since it is observed that $z=2.4848>2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>2.4848)=0.006481,$ and since $p=0.006481<0.01=\alpha,$ 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 $\alpha = 0.01$ significance level.