To calculate the test statistic for the provided data, we will conduct a paired $t$-test since each car provides paired observations for both regular and premium gas mileage. Here’s the step-by-step process:

1. Record the Data

From the table:

• Regular gas (sample 1): $16, 20, 21, 23, 22, 27, 25, 27, 28, 22$
• Premium gas (sample 2): $19, 22, 24, 25, 25, 26, 26, 28, 32, 24$

2. Calculate the Differences ($d$) Between the Two Samples

For each car: $d = \text{Premium MPG} - \text{Regular MPG}$ The differences are: $d = 3, 2, 3, 2, 3, -1, 1, 1, 4, 2$

3. Compute the Mean and Standard Deviation of Differences ($\bar{d}$ and $s_d$)

• Mean of the differences ($\bar{d}$): $\bar{d} = \frac{\sum d}{n} = \frac{(3 + 2 + 3 + 2 + 3 - 1 + 1 + 1 + 4 + 2)}{10} = \frac{20}{10} = 2$

• Standard deviation of the differences ($s_d$): $s_d = \sqrt{\frac{\sum (d - \bar{d})^2}{n - 1}}$ First, calculate $(d - \bar{d})^2$: $(3 - 2)^2 = 1, \; (2 - 2)^2 = 0, \; (3 - 2)^2 = 1, \; (2 - 2)^2 = 0, \; (3 - 2)^2 = 1, \; (-1 - 2)^2 = 9, \; (1 - 2)^2 = 1, \; (1 - 2)^2 = 1, \; (4 - 2)^2 = 4, \; (2 - 2)^2 = 0$

Sum of squared deviations: $\sum (d - \bar{d})^2 = 1 + 0 + 1 + 0 + 1 + 9 + 1 + 1 + 4 + 0 = 18$

Standard deviation: $s_d = \sqrt{\frac{18}{10 - 1}} = \sqrt{2} \approx 1.414$

4. Compute the Test Statistic ($t$-value)

The test statistic for a paired $t$-test is: $t = \frac{\bar{d}}{s_d / \sqrt{n}}$ Substitute the values: $t = \frac{2}{1.414 / \sqrt{10}} = \frac{2}{1.414 / 3.162} = \frac{2}{0.447} \approx 4.48$

Final Answer

The test statistic is approximately: $t \approx 4.48$

Would you like further assistance, such as finding the critical $t$-value or interpreting the results?

5 Related Questions:

1. How is the critical $t$-value calculated for a 90% confidence level?
2. Can you determine if the result is statistically significant?
3. What assumptions are made when conducting a paired $t$-test?
4. How does sample size affect the test statistic?
5. What is the importance of using paired differences in this scenario?

Tip:

Always check for outliers in paired data, as they can disproportionately influence the test statistic and conclusions.