Ten randomly selected cars were filled with either regular or premium gas (the decision was made based on a coin toss) and then driven on the freeway for the entirety of the tank of gas. The next day, each car was filled with whichever type of gas that was not used the previous day and then driven on the freeway for the entirety of the tank of gas. The mileage for each day was recorded and is summarized in the table below.
MPG of 10 cars (regular/premium): 13/19, 20/22, 21/24, 23/25, 22/25, 27/26, 25/26, 27/28, 28/32, 22/24.

1. The standard deviation of the differences between the two samples is what? 
2. The margin of error is what?
3. Construct the 90% confidence interval estimate mean difference between the mileage from cars using regular and premium gas.

Let’s go through each question step-by-step.

### Step 1: Calculate Differences
First, we calculate the difference in MPG for each car when using premium gas compared to regular gas. This will help us find the standard deviation of these differences.

Given data (MPG for each car with regular/premium):
- Differences (Premium - Regular): \(19 - 13 = 6\), \(22 - 20 = 2\), \(24 - 21 = 3\), \(25 - 23 = 2\), \(25 - 22 = 3\), \(26 - 27 = -1\), \(26 - 25 = 1\), \(28 - 27 = 1\), \(32 - 28 = 4\), \(24 - 22 = 2\)

So, the differences are:
\[
6, 2, 3, 2, 3, -1, 1, 1, 4, 2
\]

### Step 2: Calculate Mean and Standard Deviation of Differences
#### 1. Mean of the Differences:
\[
\text{Mean} = \frac{6 + 2 + 3 + 2 + 3 - 1 + 1 + 1 + 4 + 2}{10} = \frac{23}{10} = 2.3
\]

#### 2. Standard Deviation of the Differences:
To find the standard deviation of the differences, \( s_d \), use the formula:
\[
s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}
\]
where \( d_i \) is each difference, \( \bar{d} \) is the mean difference, and \( n \) is the sample size (10).

#### Step-by-Step Calculation:
1. Compute \( (d_i - \bar{d})^2 \) for each \( d_i \):
   - \( (6 - 2.3)^2 = 13.69 \)
   - \( (2 - 2.3)^2 = 0.09 \)
   - \( (3 - 2.3)^2 = 0.49 \)
   - \( (2 - 2.3)^2 = 0.09 \)
   - \( (3 - 2.3)^2 = 0.49 \)
   - \( (-1 - 2.3)^2 = 10.89 \)
   - \( (1 - 2.3)^2 = 1.69 \)
   - \( (1 - 2.3)^2 = 1.69 \)
   - \( (4 - 2.3)^2 = 2.89 \)
   - \( (2 - 2.3)^2 = 0.09 \)

2. Sum these values:
   \[
   \sum (d_i - \bar{d})^2 = 13.69 + 0.09 + 0.49 + 0.09 + 0.49 + 10.89 + 1.69 + 1.69 + 2.89 + 0.09 = 32.09
   \]

3. Divide by \( n - 1 \) (which is 9) and take the square root:
   \[
   s_d = \sqrt{\frac{32.09}{9}} = \sqrt{3.5656} \approx 1.89
   \]

Thus, the standard deviation of the differences is approximately **1.89**.

### Step 3: Calculate Margin of Error (90% Confidence Level)
To find the margin of error for a 90% confidence interval, use the formula:
\[
\text{Margin of Error} = t^* \cdot \frac{s_d}{\sqrt{n}}
\]
where \( t^* \) is the critical t-value for a 90% confidence interval with \( n - 1 = 9 \) degrees of freedom.

For a 90% confidence interval with 9 degrees of freedom, \( t^* \approx 1.833 \).

1. Calculate:
   \[
   \text{Margin of Error} = 1.833 \cdot \frac{1.89}{\sqrt{10}} = 1.833 \cdot 0.597 = 1.095
   \]

So, the margin of error is approximately **1.10**.

### Step 4: Construct the 90% Confidence Interval for the Mean Difference
To construct the confidence interval, use:
\[
\text{Confidence Interval} = \bar{d} \pm \text{Margin of Error}
\]
where \( \bar{d} = 2.3 \).

1. Lower bound: \( 2.3 - 1.10 = 1.2 \)
2. Upper bound: \( 2.3 + 1.10 = 3.4 \)

Therefore, the 90% confidence interval for the mean difference in mileage between regular and premium gas is **(1.2, 3.4)**.

---

### Summary of Answers:
1. **Standard deviation of the differences**: 1.89
2. **Margin of error**: 1.10
3. **90% Confidence Interval**: (1.2, 3.4)

Would you like further clarification on any of these steps?

#### Related Questions:
1. How would the confidence interval change with a 95% confidence level?
2. What would happen to the margin of error if we used a larger sample size?
3. How does changing the t-value affect the confidence interval width?
4. Can we interpret the confidence interval as the range for all future observations?
5. Why is the standard deviation calculated from the differences and not individual MPG values?

#### Tip:
When calculating confidence intervals, remember that a larger sample size narrows the margin of error, making the interval more precise.

Ten randomly selected cars were filled with either regular or premium gas (the decision was made based on a coin toss) and then driven on the freeway for the entirety of the tank of gas. The next day, each car was filled with whichever type of gas that was not used the previous day and then driven on the freeway for the entirety of the tank of gas. The mileage for each day was recorded and is summarized in the table below.
MPG of 10 cars (regular/premium): 13/19, 20/22, 21/24, 23/25, 22/25, 27/26, 25/26, 27/28, 28/32, 22/24.

1. The standard deviation of the differences between the two samples is what?
2. The margin of error is what?
3. Construct the 90% confidence interval estimate mean difference between the mileage from cars using regular and premium gas.

Analyze the 90% confidence interval for the mean difference in mileage (MPG) between regular and premium gas using sample data from 10 cars. Learn how to calculate the standard deviation of the differences, determine the margin of error, and construct a confidence interval for the mean difference. Suitable for high school statistics students.

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