Let's work through each part of the problem step by step.

Part (a): Sample Mean

You already have the sample mean $\bar{x} = 910.3$ (rounded to one decimal place), so we can move on.

Part (b): Sample Standard Deviation

The sample standard deviation $s = 306.5$ has also been provided.

Part (c): Construct a 90% Confidence Interval for the Population Mean

To calculate the confidence interval, we will use the following formula for the confidence interval of the mean when the population standard deviation is unknown:

$\text{CI} = \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$

Where:

• $\bar{x} = 910.3$ is the sample mean,
• $s = 306.5$ is the sample standard deviation,
• $n = 12$ is the sample size,
• $t_{\alpha/2}$ is the critical value from the t-distribution for a 90% confidence level with $n-1 = 11$ degrees of freedom.

Step 1: Find the critical value $t_{\alpha/2}$

For a 90% confidence interval and 11 degrees of freedom, the critical t-value can be found in a t-distribution table or using technology. The value is approximately:

$t_{\alpha/2} = 1.796$

Step 2: Calculate the margin of error (MOE)

The margin of error is given by:

$\text{MOE} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 1.796 \cdot \frac{306.5}{\sqrt{12}}$

Let me calculate the margin of error for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

For now, you can calculate the margin of error manually using the formula:

$\text{MOE} = 1.796 \cdot \frac{306.5}{\sqrt{12}} \approx 1.796 \cdot 88.45 \approx 158.8$

So, the margin of error is approximately 158.8.

Step 3: Calculate the Confidence Interval

To find the confidence interval:

$\text{Lower bound} = \bar{x} - \text{MOE} = 910.3 - 158.8 = 751.5$ $\text{Upper bound} = \bar{x} + \text{MOE} = 910.3 + 158.8 = 1069.1$

Thus, the 90% confidence interval for the population mean is approximately $(751.5, 1069.1)$.

Would you like further details or have any questions?

Here are 5 related questions for deeper exploration:

1. How does the sample size affect the width of the confidence interval?
2. Why do we use the t-distribution instead of the z-distribution for this problem?
3. What happens to the confidence interval if the sample standard deviation increases?
4. How would you interpret this confidence interval in practical terms?
5. How would a 95% confidence interval differ from a 90% one?

Tip: As the sample size increases, the t-distribution approaches the z-distribution, leading to narrower confidence intervals.