To analyze the effect of gas type on mileage, we can perform a paired t-test. Since each car was tested with both regular and premium gas, we can treat this as a paired sample (dependent) test, focusing on the difference in miles per gallon (MPG) for each car.

Step 1: Calculate Differences

For each car, find the difference in MPG when using regular gas versus premium gas: $\text{Difference (Premium - Regular)} = \text{MPG (Premium)} - \text{MPG (Regular)}$

The differences for each car are as follows:

1. $19 - 13 = 6$
2. $22 - 20 = 2$
3. $24 - 21 = 3$
4. $25 - 23 = 2$
5. $25 - 22 = 3$
6. $26 - 27 = -1$
7. $26 - 25 = 1$
8. $28 - 27 = 1$
9. $32 - 28 = 4$
10. $24 - 22 = 2$

Step 2: Calculate the Mean and Standard Deviation of Differences

Let's find the mean and standard deviation of these differences.

1. Mean of Differences: $\bar{d} = \frac{6 + 2 + 3 + 2 + 3 - 1 + 1 + 1 + 4 + 2}{10} = \frac{23}{10} = 2.3$

2. Standard Deviation of Differences: To calculate the standard deviation, we use the formula: $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}$ where $d_i$ represents each individual difference, and $n = 10$.

   Calculating each $(d_i - \bar{d})^2$:

   • $(6 - 2.3)^2 = 13.69$
   • $(2 - 2.3)^2 = 0.09$
   • $(3 - 2.3)^2 = 0.49$
   • $(2 - 2.3)^2 = 0.09$
   • $(3 - 2.3)^2 = 0.49$
   • $(-1 - 2.3)^2 = 10.89$
   • $(1 - 2.3)^2 = 1.69$
   • $(1 - 2.3)^2 = 1.69$
   • $(4 - 2.3)^2 = 2.89$
   • $(2 - 2.3)^2 = 0.09$

   Summing these: $\sum (d_i - \bar{d})^2 = 32.1$

   Now, calculate $s_d$: $s_d = \sqrt{\frac{32.1}{10 - 1}} = \sqrt{\frac{32.1}{9}} \approx 1.8856$

Step 3: Calculate the Test Statistic

The test statistic $t$ for a paired t-test is calculated as: $t = \frac{\bar{d}}{s_d / \sqrt{n}}$

Substituting the values: $t = \frac{2.3}{1.8856 / \sqrt{10}} = \frac{2.3}{0.5964} \approx 3.856$

Step 4: Determine the Degrees of Freedom and P-value

For a paired t-test, the degrees of freedom $\text{df}$ is $n - 1$, so here $\text{df} = 9$.

Using a t-distribution table or calculator, we can find the p-value associated with $t = 3.856$ and $\text{df} = 9$. This value will be relatively small, likely indicating statistical significance.

Conclusion

If the p-value is less than the chosen significance level (e.g., 0.05), we would reject the null hypothesis and conclude that there is a statistically significant difference in mileage between regular and premium gas, suggesting that premium gas may improve mileage.

Would you like the exact p-value calculation or have any further questions on this?

Related Questions

1. How would the analysis change if more cars were included in the sample?
2. What impact does the variance in mileage between cars have on the analysis?
3. Can we determine if premium gas generally provides better mileage with a larger dataset?
4. What would be the effect on the t-test if one of the mileage measurements was an outlier?
5. How does a paired t-test differ from an independent samples t-test?

Tip:

For paired tests, always calculate differences carefully, as small mistakes in differences affect the overall analysis.