Suppose that the Department of Transportation would like to test the hypothesis that the average age of cars on the road is less than 12 years. A random sample of 45 cars had an average age of 10.6 years. It is believed that the population standard deviation for the age of cars is 4.1 years. The Department of Transportation would like to set 0.05. What is the conclusion for this hypothesis test? A. Since the test statistie is more than the critical value, we can condude that the average age of cars on the road is less than 12 years. B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years. c. Since the test statistic is less than the critical value, we cannot conclude that the average age of cars on the road is less than 12 years. D. Since the test statistic is more than the critical value, we cannot conclude that the average age of cars on the road is less than 12 years.

Since our calculates values is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the best conclusion for this case is:

B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years.

Step-by-step explanation:

Data given and notation

[tex]\bar X=10.6[/tex] represent the sample mean

[tex]\sigma=4.1[/tex] represent the population standard deviation

[tex]n=45[/tex] sample size

[tex]\mu_o =12[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 is less than 12, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 12[/tex]

Alternative hypothesis:[tex]\mu < 12[/tex]

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-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:

[tex]z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29[/tex]

Critical value

For this case since w ehave a left tailed distribution we need to find a value who accumulates 0.05 of the area on th left in the normal standard distribution, and we can use the following excel code:

"=NORM.INV(0.05,0,1)"

[tex] z_{critc}= -1.64[/tex]

Conclusion

Since our calculates values is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the best conclusion for this case is:

B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years.