Answer to Question #337920 in Statistics and Probability for veve

Question

A researcher wanted to determine whether the average intelligence score for prisoners was different from the average intelligence score for the general population (μ = 100, σ = 15). Thirty-six prisoners on death row were administered an IQ test; the average IQ of the prisoners was 93. Use 0.05 significant level to test the null hypothesis.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=100$

$H_a:\mu\not=100$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{93-100}{15/\sqrt{36}}=-2.8$$

Since it is observed that $|z| = 2.8 > 1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p =2P(Z<-2.8)= 0.00511,$ and since $p = 0.00511 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

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