Question

A researcher wants to check the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such cases from court files and finds that the data was normally distributed and average jail time was 15.5 months with a standard deviation of 5.7 months. Is the reported claim correct? (use a 0.05 level of significance)

Answer 1

`t=(15.5-18.7)/((5.7)/(√(11)))=-1.862`

The degrees of freedom are given by:

`df=n-1=11-1=10`

The p value would be given by:

`p_v =2*P(t_((10))<-1.862)=0.0922`

We see that the p value is higher than the significance level so we don't have enough evidence to ocnclude that the true mean is different from 18.7 months in jail at 5% of significance.

Explanation:

Information given

`\bar X=15.5` represent the sample mean for the jail time

`s=5.7` represent the sample standard deviation

`n=11` sample size

`\mu_o =18.7` represent the value that we want to compare

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value for the test

System of hypothesis

We want to check the hypothesis if the true mean for the jail time is equal to 18.7 or no, the system of hypothesis are:

Null hypothesis:
`\mu = 18.7`

Alternative hypothesis:
`\mu \\eq 18.7`

Since we don't know the population deviation the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Rreplacing we got:

`t=(15.5-18.7)/((5.7)/(√(11)))=-1.862`

The degrees of freedom are given by:

`df=n-1=11-1=10`

The p value would be given by:

`p_v =2*P(t_((10))<-1.862)=0.0922`

We see that the p value is higher than the significance level so we don't have enough evidence to ocnclude that the true mean is different from 18.7 months in jail at 5% of significance.