To calculate the 90% confidence interval for the mean temperature, we can follow these steps:

1. Find the Sample Mean ($\bar{x}$): $\bar{x} = \frac{\sum x}{n}$

2. Calculate the Sample Standard Deviation ($s$): $s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}$

3. Determine the Margin of Error (ME): For a 90% confidence interval with $n = 12$ (small sample), we use the $t$-distribution with $n - 1 = 11$ degrees of freedom. The $t$-value for a 90% confidence level is approximately $t_{0.05, 11} = 1.796$.

   $\text{ME} = t \times \frac{s}{\sqrt{n}}$

4. Construct the Confidence Interval: The confidence interval is given by: $\left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right)$

Let's calculate each of these values.The sample mean temperature is approximately $30.87^\circ$F, with a sample standard deviation of $15.10^\circ$F.

The 90% confidence interval for the mean temperature is: $(23.04, 38.69)$

90% C.I. = $(23.04, 38.69)$

Would you like further details on the calculation?

Here are five related questions that may help deepen your understanding:

1. How would the confidence interval change with a 95% confidence level?
2. What impact would a larger sample size have on this confidence interval?
3. How do you interpret the margin of error in a confidence interval?
4. Can you use this confidence interval to make predictions about future temperatures?
5. Why do we use the $t$-distribution for small sample sizes instead of the $z$-distribution?

Tip: Increasing the confidence level (e.g., from 90% to 95%) will widen the confidence interval.