Question

In developing a new gasoline additive, researchers randomly select 10 cars and drive them both with and without the additive. The sample mean difference in gas mileage (mpg with additive - mpg without additive) is 0.41 mpg with a sample variance of 0.16. Assume the differences are from an approximately normal distribution. We want to test the hypothesis that the fuel additive has mean mpg less than the mean mpg without the additive. Calculate the test statistic.

Answer 1

Answer:

The test statistic is t = 3.24.

Step-by-step explanation:

We want to test the hypothesis that the fuel additive has mean mpg less than the mean mpg without the additive.

This means that the null hypothesis is that the difference is less than 0, while the alternate hypothesis is that the difference is 0 or more.

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $s$ is the standard deviation of the sample and n is the size of the sample.

0 is tested at the null hypothesis:

This means that $\mu = 0$

Randomly select 10 cars

This means that $n = 10$

The sample mean difference in gas mileage (mpg with additive - mpg without additive) is 0.41 mpg with a sample variance of 0.16.

This means that $X = 0.41, s = \sqrt{0.16} = 0.4$

Calculate the test statistic.

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{0.41 - 0}{\frac{0.4}{\sqrt{10}}}$

$t = 3.24$

The test statistic is t = 3.24.