Questions: 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. 1423 1228 982 690 728 724 838 625 740 546 1441 941 (a) Find the sample mean. The sample mean is 908.8 (Round to one decimal place as needed.) (b) Find the sample standard deviation. The sample standard deviation is 304 (Round to one decimal place as needed.) (c) Construct a 99% confidence interval for the population mean μ. A 99% confidence interval for the population mean is ( ) (Round to one decimal place as needed.)

Solution Steps

The sample mean \( \bar{x} \) is calculated using the formula:

\[ \bar{x} = \frac{\sum_{i=1}^N x_i}{N} \]

For the given data:

\[ \bar{x} = \frac{10906}{12} = 908.8 \]

The sample variance \( s^2 \) is calculated using the formula:

\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \]

Calculating the variance gives:

\[ s^2 = 92391.6 \]

The sample standard deviation \( s \) is then:

\[ s = \sqrt{92391.6} = 304.0 \]

To construct the confidence interval for the population mean \( \mu \) with unknown variance and small sample size, we use the formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where \( t \) is the t-value for \( n-1 \) degrees of freedom at a 99% confidence level. For \( n = 12 \), \( t \approx 3.1 \).

Calculating the margin of error:

\[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 3.1 \cdot \frac{304.0}{\sqrt{12}} \approx 272.6 \]

Thus, the confidence interval is:

\[ (908.8 - 272.6, 908.8 + 272.6) = (636.2, 1181.4) \]

Final Answer