To calculate the 90% confidence interval for the mean temperature, let's go through these steps:

1. Calculate the Sample Mean ($\bar{x}$): Add up all the sample values and divide by the number of samples.

2. Calculate the Sample Standard Deviation (s): Use the formula $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, where $x_i$ is each individual sample, $\bar{x}$ is the sample mean, and $n$ is the sample size.

3. Determine the Confidence Interval: For a 90% confidence interval with a normally distributed population and unknown population standard deviation, use the t-distribution:

   $\text{Margin of Error} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$

   where $t_{\alpha/2}$ is the t-score for a 90% confidence level with $n-1$ degrees of freedom, $s$ is the sample standard deviation, and $n$ is the sample size.

4. Calculate the Interval: The confidence interval is given by:

   $\left( \bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error} \right)$

Let’s compute this based on the data provided:

Data Provided:

Temperatures: $3.2, 36.7, 44.7, 23, 26.5, 51.7, 16.6, 44.1, 16.2, 28.4, 62.3, 75.8$

I’ll calculate the values now.The 90% confidence interval for the mean temperature, accurate to three decimal places, is:

$(24.909, 46.624)$

Would you like more details or have any questions?

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

1. How would the confidence interval change with a different confidence level, such as 95%?
2. What is the interpretation of a 90% confidence interval in this context?
3. How does the sample size affect the width of the confidence interval?
4. What is the role of the t-score, and why is it used instead of the z-score?
5. How would outliers in the data affect the confidence interval?

Tip: A larger sample size will typically result in a narrower confidence interval, assuming the sample represents the population well.