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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.

1 Answer

5 votes

Answer:


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

User Azrahel
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