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A researcher wants to know if the average time in jail for robbery has increased from what it was several years ago when the average sentence was 7 years. He obtains data on 400 recent robberies and finds an average time served of 7.5 years. If we assume the standard deviation is 3 years, perform a significance test (calculating the P-value) at the 0.05 level.

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Answer:

So, the interval is : (7.206,7.794)

Explanation:

The mean μ = 7.5

Standard deviation σ=3

n = 400

At 95% confidence interval, the z score is 1.96


7.5+1.96((3)/(√(400) ) )

And
7.5-1.96((3)/(√(400) ) )

7.5+0.294 and 7.5-0.294

So, the interval is : (7.206,7.794)

User Jason Siefken
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