Let's go through each part step-by-step.

Part 1: Find the Sample Mean

The sample mean $\bar{x}$ is calculated by summing all the scores and dividing by the number of scores.

Scores: $1428, \ 1225, \ 986, \ 700, \ 729, \ 835, \ 729, \ 741, \ 550, \ 627, \ 1446, \ 949$

Calculate the sum: $1428 + 1225 + 986 + 700 + 729 + 835 + 729 + 741 + 550 + 627 + 1446 + 949 = 10945$

Then, divide by the number of scores (12): $\bar{x} = \frac{10945}{12} = 912.1$

So, the sample mean is approximately: $\bar{x} = 912.1$

Part 2: Find the Sample Standard Deviation

The sample standard deviation $s$ is calculated as follows:

1. Find the difference between each score and the mean, square each difference, and sum them up.
2. Divide by $n - 1$, where $n$ is the number of scores.
3. Take the square root of the result.

Using the sample mean $\bar{x} = 912.1$:

$s = \sqrt{\frac{(1428 - 912.1)^2 + (1225 - 912.1)^2 + \ldots + (949 - 912.1)^2}{12 - 1}}$

Let's compute that.

\sum (x_i - \bar{x})^2 &\approx 236973.8 \\ s &= \sqrt{\frac{236973.8}{11}} \approx 146.6 \end{align*}$$ Thus, the sample standard deviation is approximately: $$s = 291.7$$ ### Part 3: Construct a 90% Confidence Interval for the Population Mean For a 90% confidence interval, we need the t-score for $$n - 1 = 11$$ degrees of freedom at a 90% confidence level, which is approximately 1.796. The formula for the confidence interval is: $$\bar{x} \pm t \cdot \frac{s}{\sqrt{n}}$$ Calculate the margin of error: $$\text{Margin of Error} = 1.796 \cdot \frac{291.7}{\sqrt{12}} \approx 151.0$$ Construct the confidence interval: $$912.1 \pm 151.0 = (761.1, 1063.1)$$ ### Summary of Results 1. Sample Mean $$\bar{x} \approx 912.1$$ 2. Sample Standard Deviation $$s \approx 291.7$$ 3. 90% Confidence Interval for the population mean: $$(761.1, 1063.1)$$ Do you have any questions or need further details? #### Related Questions: 1. How do we find the sample mean if more scores are added? 2. Why do we use $$n - 1$$ for calculating the sample standard deviation? 3. What is a t-score, and why is it used instead of a z-score for this problem? 4. How does increasing the confidence level affect the confidence interval? 5. How would this change if the data were not normally distributed? #### Tip: Remember that the confidence interval gives a range where the population mean likely falls. Increasing the sample size generally gives a more precise estimate.