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To solve this problem, we need to calculate the 90% confidence interval for the mean temperature based on the provided sample data.

### Step-by-step approach:

1. **Sample Data**:
   - Temperatures (in Fahrenheit): \( 63.2, 85.7, 46.2, 46.2, 38.8, 27.8, 51.5, 45.9 \)

2. **Calculate the Sample Mean (\(\bar{x}\))**:
   The sample mean is the sum of the data points divided by the number of data points.
   \[
   \bar{x} = \frac{\sum \text{Temperatures}}{n}
   \]

3. **Calculate the Sample Standard Deviation (\(s\))**:
   The sample standard deviation measures the spread of the data from the mean. The formula is:
   \[
   s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
   \]
   where \(x_i\) represents each individual data point.

4. **Determine the t-score**:
   Since the sample size is less than 30, we will use the t-distribution to calculate the confidence interval. For a 90% confidence interval and degrees of freedom \(n-1\), you can find the t-score from a t-distribution table.

5. **Calculate the Margin of Error (ME)**:
   The margin of error is calculated as:
   \[
   \text{ME} = t \times \frac{s}{\sqrt{n}}
   \]

6. **Compute the Confidence Interval**:
   The confidence interval is given by:
   \[
   (\bar{x} - \text{ME}, \bar{x} + \text{ME})
   \]

Let me calculate this for you.The sample mean temperature is approximately \( 50.66^\circ F \), and the sample standard deviation is \( 17.37^\circ F \).

For a 90% confidence interval, with a t-score of \( 1.895 \), the margin of error is approximately \( 11.63^\circ F \).

Thus, the 90% confidence interval for the mean temperature is:

\[
(39.03^\circ F, 62.30^\circ F)
\]

You should enter this as the open interval: \( (39.03, 62.30) \).

Let me know if you need any further details!

Here are 5 related questions to expand your understanding:
1. How does the t-distribution differ from the normal distribution in confidence intervals?
2. What would happen to the confidence interval if we increased the confidence level to 95%?
3. How would a larger sample size affect the width of the confidence interval?
4. Why is the sample standard deviation used instead of the population standard deviation?
5. What assumptions do we make when using the t-distribution for confidence intervals?

**Tip:** Always ensure that your data meets the assumptions of the statistical method you're using, such as normality when using t-distribution for small samples.

You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: [63.2, 85.7, 46.2, 46.2, 38.8, 27.8, 51.5, 45.9]. Find the 90% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places.

Learn how to calculate the 90% confidence interval for the mean temperature based on a sample dataset. The dataset consists of the temperatures: 63.2°F, 85.7°F, 46.2°F, 46.2°F, 38.8°F, 27.8°F, 51.5°F, and 45.9°F. This problem involves concepts of statistics, including the sample mean, standard deviation, and t-distribution.

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