Question

According to​ researchers, the mean length of imprisonment for​ motor-vehicle-theft offenders in a nation is 14.5 months. One hundred randomly selected​ motor-vehicle-theft offenders in a city in the nation had a mean length of imprisonment of 16.2 months.

At the 5​% significance​ level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for​ motor-vehicle-theft offenders in the city differs from the national​ mean? Assume that the population standard deviation of the lengths of imprisonment for​ motor-vehicle-theft offenders in the city is 7.0 months.

Answer 1

`z=(16.2-14.5)/((7)/(√(100)))=2.429`

The p value for this case is given by:

`p_v =2*P(z>2.429)=0.015`

The p value for this case is lower than the significance level so then we can conclude that the true mean is significantly different from 14.5 months

Explanation:

Information given

`\bar X=16.2` represent the sample mean

`\sigma=7` represent the population standard deviation

`n=100` sample size

`\mu_o =14.5` represent the value that we want to check

`\alpha=0.05` represent the significance level

z would represent the statistic

`p_v` represent the p value for the test

System of hypothesis

We want to check if the true mean is different from 14.5, the system of hypothesis would be:

Null hypothesis:
`\mu = 14.5`

Alternative hypothesis:
`\mu \\eq 14.5`

The statistic is given:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

The statistic is given by:

`z=(16.2-14.5)/((7)/(√(100)))=2.429`

The p value for this case is given by:

`p_v =2*P(z>2.429)=0.015`

The p value for this case is lower than the significance level so then we can conclude that the true mean is significantly different from 14.5 months