The gas mileages​ (in miles per​ gallon) of 24 randomly selected sports cars are listed in the accompanying table. Assume the mileages are not normally distributed. Use the standard normal distribution or the​ t-distribution to construct a 95​% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results.   20
17
 
31
21
17
22
21
23
21
20
25
23
18
22
25
21
29
20
21
nothing
21
nothing
28
nothing
25
nothing
27
nothing
21

To construct a 95% confidence interval for the mean gas mileage of the population, let's walk through the steps, given that:

1. We have a sample of 24 gas mileage values.
2. We do not know if the population is normally distributed.
3. Some values are missing in the dataset (noted as "nothing").

**Step 1: Clean the data**
We must exclude the missing values, reducing the effective sample size. After removing "nothing," the remaining values are:

\[ 20, 17, 31, 21, 17, 22, 21, 23, 21, 20, 25, 23, 18, 22, 25, 21, 29, 20, 21, 28, 25, 27, 21 \]

This leaves us with a sample size of \( n = 23 \).

**Step 2: Determine the distribution to use**
Since the sample size is less than 30 and we do not know the population standard deviation, we should use the **t-distribution** for our confidence interval. The t-distribution is more appropriate for small sample sizes and does not require the population to be normally distributed if the sample size is sufficient for the Central Limit Theorem to approximate normality.

**Step 3: Calculate sample mean and sample standard deviation**
Let's denote:
- Sample mean: \(\bar{x}\)
- Sample standard deviation: \(s\)

For the dataset provided:
\[
\bar{x} = \frac{\sum x_i}{n} \approx 22.17
\]
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \approx 3.77
\]

**Step 4: Determine the t-critical value**
For a 95% confidence level and degrees of freedom \(df = n - 1 = 22\), we find the critical value \(t_{0.025}\) (since 95% confidence leaves 2.5% in each tail):
\[
t_{0.025} \approx 2.074
\]

**Step 5: Calculate the confidence interval**
Using the formula for the confidence interval:
\[
\bar{x} \pm t_{0.025} \cdot \frac{s}{\sqrt{n}}
\]
\[
22.17 \pm 2.074 \cdot \frac{3.77}{\sqrt{23}}
\]
\[
22.17 \pm 2.074 \cdot 0.786
\]
\[
22.17 \pm 1.63
\]

So, the 95% confidence interval for the population mean is approximately:
\[
(20.54, 23.80)
\]

### Interpretation
We can be 95% confident that the true mean gas mileage for this population of sports cars lies between 20.54 and 23.80 miles per gallon.

### Summary of Solution
1. Used the t-distribution due to small sample size and unknown population standard deviation.
2. Calculated the sample mean and standard deviation.
3. Found the t-critical value for 95% confidence.
4. Computed the confidence interval.

Would you like further clarification or a breakdown of any steps?

### Further Questions
1. How does sample size impact the choice of using the t-distribution over the standard normal distribution?
2. Why is it essential to remove missing data before calculation?
3. How would the confidence interval change if we used a 99% confidence level?
4. How would using the standard normal distribution instead of the t-distribution affect this calculation?
5. What are the implications if the population distribution were known to be normal?

#### Tip:
Always check for missing values in your dataset before performing statistical calculations.

The gas mileages​ (in miles per​ gallon) of 24 randomly selected sports cars are listed in the accompanying table. Assume the mileages are not normally distributed. Use the standard normal distribution or the​ t-distribution to construct a 95​% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results.

Learn how to construct a 95% confidence interval for the mean gas mileage of a sample of sports cars using the t-distribution. This guide covers calculating the sample mean, sample standard deviation, and interpreting the results. Ideal for students in Grades 11-12 interested in statistical inference.

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