Answer to Question #339699 in Statistics and Probability for Agl

Question

3. A psychologist believes that it will take at least an hour for certain disturbed

children to learn a task. A random sample of 30 of these children results in a

mean of 50 minutes to learn the task. Should the psychologist modify her belief

at the 0.01 level if the population standard deviation can be assumed to be 15

minutes?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge 60$

$H_a:\mu<60$

This corresponds to a left-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 left-tailed test is $z_c = -2.3263.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{50-60}{15/\sqrt{30}}\approx-3.6515$$

Since it is observed that $z = -3.6515 <-2.3263= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<-3.6515)=0.00013,$ and since $p=0.00013<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 60, at the $\alpha = 0.01$ significance level.