QUESTION 1

Find the confidence interval specified. Assume that the population is normally distributed.
Thirty randomly selected students took the calculus final. If the sample mean was 81 and the standard deviation was 9.3, construct a $99\mathrm{\%}$ confidence interval for the mean score of all students.
78.12 to 83.88
76.34 to 85.66
76.82 to 85.18
76.32 to 85.68
6.67 points
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Step 1 ：To construct a 99% confidence interval for the mean score of all students, the formula is $\text{CI}=\overline{x}\pm z\times \left(\frac{s}{\sqrt{n}}\right)$, where $\overline{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level, $s$ is the sample standard deviation, and $n$ is the sample size.

Step 2 ：The sample mean $\overline{x}$ is 81, the sample standard deviation $s$ is 9.3, and the sample size $n$ is 30.

Step 3 ：For a 99% confidence level, the z-score $z$ is approximately 2.576.

Step 4 ：The margin of error $E$ is calculated as $E=z\times \frac{s}{\sqrt{n}}$.

Step 5 ：Substituting the known values gives $E=2.576\times \frac{9.3}{\sqrt{30}}\approx 4.37$.

Step 6 ：The confidence interval is then $\text{CI}=81\pm 4.37$.

Step 7 ：Calculating the lower and upper bounds gives $81-4.37=76.63$ and $81+4.37=85.37$.

Step 8 ：The confidence interval is $(76.63,85.37)$.

Step 9 ：Rounding to two decimal places, the confidence interval is $(76.63,85.37)$.

Step 10 ：Comparing with the given options, the closest is $76.32\text{ to }85.68$ when rounded.

Step 11 ：The correct confidence interval is $\boxed{{\displaystyle 76.63\text{ to }85.37}}$.