Answer:
The test statistic value is -0.92.
Step-by-step explanation:
In this case we need to test whether the van has an incorrect manufacturer's MPG rating.
The hypothesis can be defined as follows:
<em>H₀</em>: The manufacturer's MPG rating for the van is 54.5, i.e. <em>μ</em> = 54.5.
<em>Hₐ</em>: The manufacturer's MPG rating for the van is 54.5, i.e. <em>μ</em> ≠ 54.5.
The information provided is:
![n=220\\\bar x=54.2\\\sigma=4.84\\\alpha =0.01](https://tex.z-dn.net/?f=n%3D220%5C%5C%5Cbar%20x%3D54.2%5C%5C%5Csigma%3D4.84%5C%5C%5Calpha%20%3D0.01)
As the population standard deviation is provided, we will use a z-test for single mean.
Compute the test statistic value as follows:
![z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{54.2-54.5}{4.84/\sqrt{220}}=-0.92](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Cbar%20x-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B54.2-54.5%7D%7B4.84%2F%5Csqrt%7B220%7D%7D%3D-0.92)
The test statistic value is -0.92.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected.
Compute the <em>p</em>-value for the two-tailed test as follows:
![p-value=2\ P(Z](https://tex.z-dn.net/?f=p-value%3D2%5C%20P%28Z%3C-0.92%29%5C%5C%3D2%5Ctimes%200.17879%5C%5C%3D0.35758%5C%5C%5Capprox%200.36)
*Use a z-table for the probability.
The p-value of the test is 0.36.
The p-value of the test is very large.
<em>p</em>-value = 0.36 > α = 0.01
The null hypothesis will not be rejected.
Thus, it can be concluded that the manufacturer's MPG rating for the van is 54.5 miles/gallon.