The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 2014. Being from Pennsylvania at that time, you believed automobile insurance was cheaper there and decided to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1,430 with a standard deviation of s = $160.(a)Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium.H0: μ > 1,503Ha: μ ≤ 1,503H0: μ ≤ 1,503Ha: μ > 1,503    H0: μ = 1,503Ha: μ ≠ 1,503H0: μ < 1,503Ha: μ ≥ 1,503H0: μ ≥ 1,503Ha: μ < 1,503(b)What is a point estimate in dollars of the difference between the mean annual premium in Pennsylvania and the national mean? (Use the mean annual premium in Pennsylvania minus the national mean.)$ (c)At α = 0.05, test for a significant difference.Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.)p-value =

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Asked Dec 12, 2019
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The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 2014. Being from Pennsylvania at that time, you believed automobile insurance was cheaper there and decided to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1,430 with a standard deviation of 
s = $160.
(a)
Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium.
H0: μ > 1,503
Ha: μ ≤ 1,503
H0: μ ≤ 1,503
Ha: μ > 1,503
    
H0: μ = 1,503
Ha: μ ≠ 1,503
H0: μ < 1,503
Ha: μ ≥ 1,503
H0: μ ≥ 1,503
Ha: μ < 1,503
(b)
What is a point estimate in dollars of the difference between the mean annual premium in Pennsylvania and the national mean? (Use the mean annual premium in Pennsylvania minus the national mean.)
(c)
At 
α = 0.05,
 test for a significant difference.
Find the value of the test statistic. (Round your answer to three decimal places.)
 
Find the p-value. (Round your answer to four decimal places.)
p-value = 
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Expert Answer

Step 1

Null hypothesis:

µ=1,503.

Alternative hypothesis:

µ<1,503.

This is a left tailed test.

Since the population standard deviation is unknown, the appropriate test is one sample t-test.

Here, the sample mean, x-bar is 1,430.

Population mean, µ is 1,503.

Sample standard deviation, s = 160.

Sample size, n is 25.

The test statistic value can be obtained as follows:

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t = In 1,430 – 1,503 160 25 x -2.28

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Step 2

Computation of P-value:

The P-value t-distribution at 24 degrees of freedom can be obtained using the exc...

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fe =T.DIST(-2.28,24,TRUE) 0.015892527

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Hypothesis Testing

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