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

Answer:

$z=\frac{16.2-14.5}{\frac{7}{\sqrt{100}}}=2.429$    

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

Step-by-step 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:$\mu \neq 14.5$  

The statistic is given:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

The statistic is given by:

$z=\frac{16.2-14.5}{\frac{7}{\sqrt{100}}}=2.429$    

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