Question #204204

Perpetrators of a crime who have been given the maximum sentence of 10 years are believed by a fiscal to have been sentenced to no more than 7 years on average. A random sample of 20 of 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 0.05 level of significance?


1
Expert's answer
2021-06-13T17:28:39-0400

μ=7n=20xˉ=7.6s=2.4α=0.05H0:μ=7H1:μ>7df=n1=201=19\mu=7 \\ n=20 \\ \bar{x}=7.6 \\ s=2.4 \\ α=0.05 \\ H_0: \mu=7 \\ H_1: \mu > 7 \\ df=n-1 = 20-1=19

Test statistic:

t=xˉμs/n=7.672.4/20=0.60.5366=1.118t= \frac{\bar{x}- \mu}{s / \sqrt{n}} \\ = \frac{7.6-7}{2.4 / \sqrt{20} } \\ = \frac{0.6}{0.5366} \\ = 1.118

Read the t-table value at 5% level of significance at 19 degrees of freedom for Right tailed test is 1.72

t-cal<t-tab

So, we accept H0 and reject H1.

We can conclude, that the fiscal is correct in his suspicions at 0.05 level of significance.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS