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.

Question

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.