Question

An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: 27.2 \(\begin{array}{lllll}29.3 & 31.2 & 28.4 & 30.3 & 29.6\end{array}\) Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) \(30 \mathrm{mpg}\) ?

Short answer

To complete this analysis, perform the calculations as described in steps 1-6. If the calculated t-value is less than the critical t-value, then there is not enough evidence to refute the claim that actual mean fuel efficiency for this model is at least 30 mpg. Otherwise, the claim is not supported by the data.

Step-by-step solution

Organize the Data

Firstly, list out the given fuel efficiencies for the six different people: 27.2, 29.3, 31.2, 28.4, 30.3, 29.6.

Calculate the Sample Mean

To find the sample mean, add up all these fuel efficiencies, and then divide by the number of observations, which in this case is 6. So, the sample mean is \(\frac{27.2+29.3+31.2+28.4+30.3+29.6}{6}\).

Calculate the Sample Standard Deviation

To find the sample standard deviation, first find the squared difference between each observation and the sample mean, add those squared differences, divide by the number of observations minus 1, and then take the square root of the result.

Calculate the t-value

Next, calculate the t-value. This is calculated as \(t = \frac{sample \ mean - µ0}{sample \ standard \ deviation / \sqrt{n}}\). Here, µ0 = 30 (the assumed population mean), n = 6 (the sample size).

Determine the critical t-value

Look up the critical value for a one-tailed test (since we're interested in whether the actual mean is less than 30 mpg) from a t-table. Use a significance level of 0.05 and degrees of freedom = 5 (n - 1). This will be the critical t-value.

Compare the t-values

If the calculated t-value is less than the critical t-value, then we would accept the null hypothesis. Otherwise, we reject the null hypothesis.