Question

Assume that a sample is used to estimate a population mean mu. Find the 99% confidence interval for a sample of size 45 with a mean of 78.2 and a standard deviation of 10.7. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

99% C.I. =

Answer 1

To calculate the 99% confidence interval for the population mean \(\mu\), we can use the formula for the confidence interval based on the sample mean, the critical value from the standard normal distribution (Z-distribution), and the standard error of the mean.

1. **Calculate the standard error (SE)**:
\[
SE = \frac{s}{\sqrt{n}}
\]
where \(s\) is the sample standard deviation, and \(n\) is the sample size.

Given:
- \(s = 10.7\)
- \(n = 45\)

\[
SE = \frac{10.7}{\sqrt{45}} \approx \frac{10.7}{6.708} \approx 1.592
\]

2. **Find the critical value for a 99% confidence level**:
For a two-tailed test at a 99% confidence level, we look up the Z-score that corresponds to an area of \(0.005\) in each tail (since \(0.01/2 = 0.005\)). This critical value is approximately \(Z \approx 2.576\).

3. **Calculate the margin of error (ME)**:
\[
ME = Z \times SE
\]
\[
ME = 2.576 \times 1.592 \approx 4.102
\]

4. **Construct the confidence interval**:
The confidence interval can be calculated using the sample mean and the margin of error.
\[
\text{Confidence Interval} = (\bar{x} - ME, \bar{x} + ME)
\]
Where \(\bar{x} = 78.2\).

\[
\text{Lower limit} = 78.2 - 4.102 \approx 74.1
\]
\[
\text{Upper limit} = 78.2 + 4.102 \approx 82.3
\]

Thus, the 99% confidence interval for the population mean \(\mu\) is:

\[
\text{99% C.I.} = (74.1, 82.3)
\]