Question

The mean annual premium for automobile insurance in the United States is $1503 (Insure website, March 6, 2014). Being from Pennsylvania, you believe automobile insurance is cheaper there and wish 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 $1440 with a standard deviation of s = $165. If required, enter negative values as negative numbers.
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.

Answer 1

Answer:

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

$\bar X=1440$ represent the mean annual premium value for the sample  

$s=165$ represent the sample standard deviation for the sample  

$n=25$ sample size  

$\mu_o =1503$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean annual premium in Pennsylvania is lower than the national mean annual premium, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 1503$  

Alternative hypothesis:$\mu < 1503$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{1440-1503}{\frac{165}{\sqrt{25}}}=-1.909$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=25-1=24$  

Since is a one side left tailed test the p value would be:  

$p_v =P(t_{(24)}$  

Conclusion  

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.