Question

Imagine an automobile company looking for additives that might increase gas mileage. As a pilot study, they send 30 cars fueled with a new additive on a road trip from Boston to Los Angeles. Without the additive, those cars are known to average 25.0mpg with a standard deviation of 2.4 mpg. Suppose it turns out that the thirty cars averaged 26.3 mpg with the additive. What should the company conclude? Is the additive effective? Let α=0.01.

a)Use three methods: the p-value, the critical value approach and the confidence interval method.

b) Describe what a type I error would be. Describe what a type II error would be.

Answer 1

(a)

Hypotheses are

:

Test is right tailed.

Given information:

Since population standard deviation is known so single sample z test will be used.

The test statistics is

Critical value approach:

Test is right tailed so using excel function "=NORMSINV(0.99)" the critical value of z is 2.33. (z-table can also be used).

The rejection region:

If z > 2.33, reject H0

Decsion:

Since test statistics lies in the rejection region so we reject the null hypothesis.

p-value approach:

Using excel function "=1-NORMSDIST(2.97)" the p-value of the test is

p-value = 0.0015

Decsion:

Since p-value is less than 0.01 so we reject the null hypothesis. Since test statistics lies in the rejection region so we reject the null hypothesis.

Confidence interval:

For 99% confidence interval, using excel function "=NORMSINV(0.995)", critical value of z is . So required confidence interval is

Therefore, a 99% confidence interval for the mean is (25.172, 27.428).

Decision: Since 25 mpg is not lie in the above interval so we reject the null hypothesis on the basis of confidence interval. Since all values are greater than 25 mpg so we can conclude that the additive is effective.

b)

Type I error: It is probability of rejecting the true null hypothesis.

That is we incorrectly conclude that  the additive effective while actually it is not.

Type II error: It is probability of fail to reject the true null hypothesis.

That is we incorrectly conclude that the additive effective is not effective while actually it is.