The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. 1421 1230 981 697 726 839 721 744 541 627 1447 946 (a) Find the sample mean. $\bar{x}=910.0$ (Round to one decimal place as needed.) (b) Find the sample standard deviation. $s=304.8$ (Round to one decimal place as needed.) (c) Construct a $90 \%$ confidence interval for the population mean $\mu$. A $90 \%$ confidence interval for the population mean is

Step 1 ：Given the state test scores for 12 randomly selected high school seniors are 1421, 1230, 981, 697, 726, 839, 721, 744, 541, 627, 1447, 946.

Step 2 ：(a) To find the sample mean, we sum up all the scores and divide by the number of scores. The sample mean \(\bar{x}\) is calculated as \(\frac{1421+1230+981+697+726+839+721+744+541+627+1447+946}{12} = 910.0\).

Step 3 ：(b) To find the sample standard deviation, we first find the variance, which is the average of the squared differences from the mean. Then, we take the square root of the variance to get the standard deviation. The sample standard deviation \(s\) is calculated as \(\sqrt{\frac{(1421-910)^2+(1230-910)^2+(981-910)^2+(697-910)^2+(726-910)^2+(839-910)^2+(721-910)^2+(744-910)^2+(541-910)^2+(627-910)^2+(1447-910)^2+(946-910)^2}{12-1}} = 304.8\).

Step 4 ：(c) To construct a 90% confidence interval for the population mean, we use the formula \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level (for a 90% confidence interval, \(z\) is approximately 1.645), \(s\) is the sample standard deviation, and \(n\) is the sample size. The 90% confidence interval for the population mean is \(910 \pm 1.645 \frac{304.8}{\sqrt{12}} = (765.3, 1054.7)\).

Step 5 ：Final Answer: The sample mean is \(\boxed{910.0}\), the sample standard deviation is \(\boxed{304.8}\), and a 90% confidence interval for the population mean is \(\boxed{(765.3, 1054.7)}\).