Site_A	Site_B
$115	$110
$79	$79
$43	$40
$140	$129
$97	$97
$33	$33
$84	$69
$95	$95
$118	$108
$69	$66
c. Using software or a​ calculator, construct a​ 90% confidence interval comparing the population mean prices of all textbooks used that term at her college​ (price at Site A minus price at Site​ B). Interpret.

To construct a 90% confidence interval for the difference in population means (price at Site A minus price at Site B), we can follow these steps:

### 1. **Identify the Data**
Here are the prices from Site A and Site B:

\[
\begin{array}{c|c|c}
\text{Site A} & \text{Site B} & \text{Difference (A - B)} \\
\hline
115 & 110 & 5 \\
79  & 79  & 0 \\
43  & 40  & 3 \\
140 & 129 & 11 \\
97  & 97  & 0 \\
33  & 33  & 0 \\
84  & 69  & 15 \\
95  & 95  & 0 \\
118 & 108 & 10 \\
69  & 66  & 3 \\
\end{array}
\]

### 2. **Calculate the Mean and Standard Deviation of Differences**
Let's calculate the mean and standard deviation of the differences.

\[
\text{Differences} = \{5, 0, 3, 11, 0, 0, 15, 0, 10, 3\}
\]

### 3. **Calculate the Mean Difference ($\bar{d}$)**

\[
\bar{d} = \frac{5 + 0 + 3 + 11 + 0 + 0 + 15 + 0 + 10 + 3}{10} = \frac{47}{10} = 4.7
\]

### 4. **Calculate the Standard Deviation of Differences ($s_d$)**

First, find the squared differences from the mean for each difference:

\[
(5 - 4.7)^2 = 0.09, \, (0 - 4.7)^2 = 22.09, \, (3 - 4.7)^2 = 2.89, \, \dots
\]

Then, find the sum of these squared differences, divide by $n-1$ (where $n=10$), and take the square root to find $s_d$.

\[
s_d = \sqrt{\frac{1}{9} \sum_{i=1}^{10} (d_i - \bar{d})^2}
\]

Let's calculate this:

### 5. **Calculate the 90% Confidence Interval**

For a 90% confidence interval with $n=10$, the critical t-value ($t_{\alpha/2}$) can be found from the t-distribution table with 9 degrees of freedom.

The confidence interval is given by:

\[
\bar{d} \pm t_{\alpha/2} \cdot \frac{s_d}{\sqrt{n}}
\]

### 6. **Interpret the Confidence Interval**

Let's calculate these values and then interpret the result.### Confidence Interval Calculation Results:

- **Mean Difference ($\bar{d}$)**: $4.7$
- **Standard Deviation of Differences ($s_d$)**: $5.46$
- **t-Critical Value for 90% Confidence Level**: $1.83$
- **Margin of Error**: $3.16$
- **90% Confidence Interval**: $(1.54, 7.86)$

### Interpretation:
We are 90% confident that the true mean difference in textbook prices between Site A and Site B (Site A minus Site B) is between $1.54$ and $7.86$. This suggests that, on average, textbooks at Site A are likely to be priced higher than those at Site B by an amount within this range.

### Do you have any questions, or would you like further details?

---

Here are some additional related questions:
1. How is the t-critical value determined for a confidence interval?
2. What would the confidence interval look like for a 95% confidence level?
3. How does increasing the sample size affect the confidence interval?
4. What assumptions are made when constructing a confidence interval for the mean difference?
5. What is the interpretation of the margin of error in this context?
6. How would you calculate a confidence interval using a calculator?
7. What does it mean if the confidence interval includes zero?
8. How do outliers in the data affect the confidence interval?

**Tip:** Always check the assumptions of normality and equal variances before using a t-distribution to construct a confidence interval for the difference between two means.

Learn how to construct a 90% confidence interval for the difference in textbook prices between Site A and Site B. This detailed solution uses statistical methods including the t-distribution and provides a step-by-step calculation of the confidence interval.

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