Question

sample of size n=39 has sample mean x=55 and sample standard deviation s=5.8. (a) Construct a 95% confidence interval for the population mean u. Enter the values for the lower and upper limits and the mean to graph. Round the answers to one decimal place.

Answer 1

To construct a 95% confidence interval for the population mean \( \mu \), we can use the formula:

\[
\text{CI} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]

where:
- \( \bar{x} \) is the sample mean,
- \( t^* \) is the critical value from the t-distribution for \( n-1 \) degrees of freedom,
- \( s \) is the sample standard deviation,
- \( n \) is the sample size.

Given:
- Sample size \( n = 39 \)
- Sample mean \( \bar{x} = 55 \)
- Sample standard deviation \( s = 5.8 \)

1. **Calculate degrees of freedom**:
\[
df = n - 1 = 39 - 1 = 38
\]

2. **Determine the critical value \( t^* \)** for a 95% confidence level and 38 degrees of freedom. You can find this from a t-table or using software. The critical value \( t^* \approx 2.024 \).

3. **Calculate the standard error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{5.8}{\sqrt{39}} \approx \frac{5.8}{6.244} \approx 0.928
\]

4. **Calculate the margin of error (ME)**:
\[
ME = t^* \cdot SE \approx 2.024 \cdot 0.928 \approx 1.876
\]

5. **Construct the confidence interval**:
\[
\text{Lower limit} = \bar{x} - ME = 55 - 1.876 \approx 53.1
\]
\[
\text{Upper limit} = \bar{x} + ME = 55 + 1.876 \approx 56.9
\]

Thus, the 95% confidence interval for the population mean \( \mu \) is approximately:

- **Lower limit**: 53.1
- **Upper limit**: 56.9

Finally, to summarize the answer:
- Lower limit: **53.1**
- Upper limit: **56.9**
- Mean: **55**