An automobile manufacturer has given its van a 59.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating:

At the null hypothesis, we test if the mean is the same, that is:

$H_0: \mu = 59.5$

At the alternate hypothesis, we test that it is different, that is:

$H_a: \mu \neq 59.5$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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

59.5 is tested at the null hypothesis:

This means that $\mu = 59.5$

After testing 250 vans, they found a mean MPG of 59.2. Assume the population standard deviation is known to be 1.9.

This means that $n = 250, X = 59.2, \sigma = 1.9$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{59.2 - 59.5}{\frac{1.9}{\sqrt{250}}}$

$z = -2.5$

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean that differs from 59.5 by at least 0.3, which is P(|Z|>-2.5), which is 2 multiplied by the pvalue of Z = -2.5.

Looking at the z-table, Z = -2.5 has a pvalue of 0.0062

2*0.0062 = 0.0124

The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

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3 years ago