Question

A recent report states the variance of the fuel consumption of a certain car is equal to 16.81 miles per gallon (mpg). A researcher claims the actual variance is lower. A sample of 21 cars produced a variance of 6.25 mpg. Test the claim at alpha = 0.05, use the P-value method. For question #29, tell what type of test his is (right, left, or two-tailed). For question #30, select from choices given with regard to identifying the alternate hypothesis. For question #31, fill in the blank with your Critical Value(s) for Chi-square. For question #32, fill in the space provided with your calculated value of Chi-square. For question #33, use your answer from #32 to select between which two areas your P-value lies. For question #34, based on your answer for #33, answer True or False as to whether the null hypothesis will be rejected.

Answer 1

Answer:

H0: $\sigma \geq 4.1$

H1: $\sigma <4.1$

$\Chi^2_{crit} =10.851$

$t=(21-1) [\frac{2.5}{4.1}]^2 =7.436$

$p_v = P(\Chi^2_{20}<7.436)=0.0050$

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

$\bar X$ represent the sample mean

n = 21 sample size

$s^2 =6.25$ represent the sample variance

$s=2.5$ represent the sample deviation

$\sigma^2_0 =16.81$ represent the target variance

$\sigma_o =4.1$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic

$\alpha=0.05$ significance level

State the null and alternative hypothesis

On this case we want to check if the population variance is lower (or if the deviation is lower than 4.1) than 16.81, so the system of hypothesis are:

H0: $\sigma \geq 4.1$

H1: $\sigma <4.1$

In order to check the hypothesis we need to calculate th statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a left tailed test the critical zone it's on the left tail of the distribution. On this case we need a quantile on the chi square distribution with 20 degrees of freedom that accumulates 0.05 of the area on the left tail and 0.95 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.05,20)". And our critical value would be $\Chi^2_{crit} =10.851$

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(21-1) [\frac{2.5}{4.1}]^2 =7.436$

What is the approximate p-value of the test?

For this case since we have a left tailed test the p value is given by:

$p_v = P(\Chi^2_{20}<7.436)=0.0050$

Based on your answer for the p value, answer True or False as to whether the null hypothesis will be rejected.

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)