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3. A report stated that the 95% confidence interval for the average gas mileage in miles based on 49 SUVS is (18, 22). The re

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Answer #1

#1.
ME = (22 - 18)/2 = 2

ME = z*sigma/sqrt(n)
z = 1.96
n = 49

sigma = 2*7/1.96 = 7.1429

#2.
xbar = (18 + 22)/2 = 40/2 = 20

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 17
Alternative Hypothesis, Ha: μ > 17

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (20 - 17)/(7.1429/sqrt(49))
z = 2.94

P-value Approach
P-value = 0.0016
As P-value < 0.01, reject the null hypothesis.

Yes, mileage is greater than 17 miles

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