Answer to Question #211161 in Statistics and Probability for tel

Question

A psychologist believes that it will take at least an hour for certain disturbed children to learn a task. A

random sample of 35 of these children results in a mean of 50 minutes to learn the task. Should the

psychologist modify her belief at 0.01 level if the population standard deviation can be assumed to be 15

minutes.

Answer 1

Solution:

Hypothesized Population Mean $\mu=60\ min$

Standard Deviation $\sigma=15\ min$

Sample Size $n=30$

Sample Mean $\bar{x}=50\ min$

Significance Level $\alpha=0.01$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq60$

$H_1: \mu<60$

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

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

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.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than $60,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for left-tailed, the significance level $\alpha=0.01, z=-3.6515,$ is $p=0.00013,$ 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.