Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.

A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the mean score of all such subjects.

A) 67.7 < μ < 84.7
B) 69.2 < μ < 83.2
C) 74.6 < μ < 77.8
D) 64.2 < μ < 88.2

Answer 1

To construct a 95% confidence interval for the population mean μ based on the sample data provided, we can follow these steps:

1. **Identify the sample statistics:**
- Sample mean (\( \bar{x} \)) = 76.2
- Sample standard deviation (s) = 21.4
- Sample size (n) = 27

2. **Determine the critical value:**
Since the sample size is small (n < 30), we will use the t-distribution to find the critical value. For a 95% confidence interval and a sample size of 27, the degrees of freedom (df) will be \( n - 1 = 27 - 1 = 26 \).

Using a t-table or calculator, the critical t-value (t*) for a 95% confidence interval with 26 degrees of freedom is approximately 2.056.

3. **Calculate the standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = \frac{21.4}{\sqrt{27}} \approx \frac{21.4}{5.196} \approx 4.12
\]

4. **Calculate the margin of error (ME):**
\[
ME = t^* \times SE = 2.056 \times 4.12 \approx 8.465
\]

5. **Construct the confidence interval:**
\[
CI = \bar{x} \pm ME = 76.2 \pm 8.465
\]
This results in:
- Lower limit: \( 76.2 - 8.465 \approx 67.735 \)
- Upper limit: \( 76.2 + 8.465 \approx 84.665 \)

Thus, rounding to one decimal place, the 95% confidence interval is approximately:

\[
67.7 < \mu < 84.7
\]

So the correct answer is **A) 67.7 < μ < 84.7**.