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Dovator [93]
3 years ago
11

According to​ researchers, the mean length of imprisonment for​ motor-vehicle-theft offenders in a nation is 14.5 months. One hu

ndred 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.
Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

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

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