Answer to Question #339334 in Statistics and Probability for diane

Question

Perpetrators of a crime who have been given the maximum sentence of 10 years are

believed bya fiscal to have been sentenced to no more than 7 years on the average.

random sample of 20of these crime offenders reveals a mean sentence of 7.6 years with a

standard deviation of 2.4 years. Is the fiscal correct in his suspicions at the 0.05 level of

significance?

Step:

1. State the null and alternative hypothesis

concerning the population mean, $\mu$ and the type of test to be used.

2.Specify the level of significance $\alpha$

3.State the decision rule

4. Collect the data and perform calculations.

5. Make a statistical decision.

6.State the conclusion

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu\le7$

$H_a:\mu>7$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

2. Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=19$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 1.729133.$

3. The rejection region for this right-tailed test is $R = \{t: t> 1.729133\}.$

4. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{7.6-7}{2.4/\sqrt{20}}\approx1.118$$

5. Since it is observed that $t=1.118< 1.729133=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=19$ degrees of freedom, $t=1.118$ is $p=0.138752,$ and since $p=0.138752>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

6. Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 7, at the $\alpha = 0.05$ significance level.