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SVETLANKA909090
10 days ago
8

The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 20

14. 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.
Mathematics
1 answer:
PIT_PIT [9.1K]10 days ago
8 0

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. This indicates that the average automobile premium in Pennsylvania is lower than in the nation.

Step-by-step breakdown:

This is a statistical test for the average population mean.

The hypothesis posits that car insurance in Pennsylvania is notably less expensive compared to the national average.

Accordingly, the null and alternative hypotheses are:

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

The significance level is set at 0.05.

The sample size is n=25.

The sample mean equates to M=1425.

A point estimate of the difference between the Pennsylvania mean premium and the national average can be computed using the sample mean:

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

Given that the standard deviation of the population is unknown, we approximate it using the sample standard deviation, which is s=160.

The estimated standard error of the mean is determined with 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 stand at:

df=n-1=25-1=24

This constitutes a left-tailed test, with 24 degrees of freedom and t=-2.438, rendering the P-value as (per a t-table):

\text{P-value}=P(t

As the P-value (0.0113) falls below the significance level (0.05), the results prove significant.

Thus, the null hypothesis obtains dismissal.

At a 0.05 significance level, there's sufficient evidence to assert that car insurance in Pennsylvania costs notably less than the national average.

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Explanation

This can be simplified as follows:

  • 10 times a given quantity of _____ hundreds results in 60 hundreds
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We will divide this into two segments:

\boxed{ \ 10 \times (N \times 100) = 60 \times 100 \ }... Equation-1

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This straightforward multiplication relates to the place value system. We need to perform calculations to find the values of N and M.

Initially, let's focus on Equation-1.

We will move the variable M to one side of the equation in order to isolate it and solve for its value.

\boxed{ \ 10 \times (N \times 100) = 60 \times 100 \ }

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Both parts will be divided by 1,000, essentially being multiplied by \frac{1}{1,000}.

\boxed{ \ \frac{1}{1,000} \times 1,000 \times N = \frac{1}{1,000} \times 6,000 \ }

Thus, we conclude with \boxed{\boxed{ \ N = 6 \ }}

Next, we process Equation-2 to derive M's value.

\boxed{ \ 60 \times 100) = M \times 1,000 \ }

\boxed{ \ 6,000 = M \times 1,000 \ }

Rearranging this equation to place M on the left side appears as follows:

\boxed{ \ M \times 1,000 = 6,000 \ }

Again, both sides undergo division by 1,000, which translates to multiplication by \frac{1}{1,000}..

\boxed{ \ \frac{1}{1,000} \times M \times 1,000 = \frac{1}{1,000} \times 6,000 \ }

This results in \boxed{\boxed{ \ M = 6 \ }}

- - - - - - -

Alternative approach for the second step:

60 hundreds equals to __ thousands

\boxed{ \ 6 \times 10 \ hundreds = \ M \ thousands} \ }

\boxed{ \ 6 \times thousands = \ M \ thousands} \ }

\boxed{\boxed{ \ M = 6 \ }}

Additional resources

  1. A more detailed version of this topic
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  3. 100 is equivalent to 1/10 of which number?

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