Question

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

2. According to the norms established for a history test, grade 8 students should have an average of 81.7 with a standard deviation of 8.5.

Answer 1

1. A psychologist believes that it will take at least an hour for a 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 then psychologist modify her belief at the 0.01 level if the population standard deviation can be assumed to be 15 minutes?

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.

2. According to the norms established for a history test, grade 8 students should have an average of 81.7 with a standard deviation of 8.5. If 100 randomly selected grade 8 students from a certain school district average 79.6 in this test, can we conclude at the 0.05 level of significance that the grade 8 students from this school district can be expected to average less than the norm of 81.7?

Hypothesized Population Mean $\mu=81.7$

Standard Deviation $\sigma=8.5$

Sample Size $n=100$

Sample Mean $\bar{x}=79.6$

Significance Level $\alpha=0.05$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq81.7$

$H_1: \mu<81.7$

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 $\alpha=0.05,$and the critical value for left-tailed test is $z_c=-1.6449.$

The rejection region for this left-tailed test is $R=\{t:z<-1.6449\}.$

The $z$ - statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{79.6-81.7}{8.5/\sqrt{100}}\approx-2.4706$$

Since it is observed that $z=-2.4706<-1.6449=z_c,$ 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 $81.7,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for left-tailed, the significance level $\alpha=0.01, z=-2.4706,$ is $p=0.00674,$ and since $p=0.00674<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than $81.7,$ at the $\alpha=0.05$ significance level.