Question

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 $1,425 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 =
State your conclusion.
Do not reject H0. We can conclude that the population mean automobile premium in Pennsylvania is lower than the national mean.Reject H0. We cannot conclude that the population mean automobile premium in Pennsylvania is lower than the national mean. Do not reject H0. We cannot conclude that the population mean automobile premium in Pennsylvania is lower than the national mean.Reject H0. We can conclude that the population mean automobile premium in Pennsylvania is lower than the national mean.

Answer 1

Answer:

a) Null and alternative hypothesis

$H_0: \mu=1503\\\\H_a:\mu< 1503$

b)  Point estimate d = -$78

c) Test statistic t = -2.438

P-value = 0.0113

Reject H0. We can conclude that the population mean automobile premium in Pennsylvania is lower than the national mean.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that automobile insurance in Pennsylvania is significantly cheaper than the national average.

Then, the null and alternative hypothesis are:

$H_0: \mu=1503\\\\H_a:\mu< 1503$

The significance level is 0.05.

The sample has a size n=25.

The sample mean is M=1425.

A point estimate of the difference between the mean annual premium in Pennsylvania and the national mean can be calculated with the sample mean:

$d=M-\mu=1425-1503=-78$

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=160.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{160}{\sqrt{25}}=32$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1425-1503}{32}=\dfrac{-78}{32}=-2.438$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

This test is a left-tailed test, with 24 degrees of freedom and t=-2.438, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0113) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that automobile insurance in Pennsylvania is significantly cheaper than the national average.