Question

in a random sample of 18 people, the mean commute time to work was 34.7 minutes and the standard deviation was 7.1 minutes. assume the population is normally distributed to construct a 80% confidence interval for the population mean μ. what is the margin of error of μ? interpret the results. the confidence interval for the population mean μ is (round to one decimal place as needed)

Explanation:

Step1: Identify the distribution and formula

We have a sample size \( n = 18 \), sample mean \( \bar{x}=34.7 \), sample standard deviation \( s = 7.1 \). Since the population standard deviation is unknown and \( n<30 \), we use the t - distribution. The formula for the margin of error \( E \) for a confidence interval of the population mean (when using t - distribution) is \( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}} \), and for the confidence interval, it is \( \bar{x}\pm E \). For an 80% confidence interval, \( \alpha=1 - 0.8=0.2 \), so \( \alpha/2 = 0.1 \). The degrees of freedom \( df=n - 1=18 - 1 = 17 \).

Step2: Find the t - critical value

Looking up the t - distribution table for \( df = 17 \) and \( \alpha/2=0.1 \), we find that \( t_{0.1,17}\approx1.333 \) (using t - table or statistical software).

Step3: Calculate the margin of error

Substitute \( t_{\alpha/2}=1.333 \), \( s = 7.1 \), and \( n = 18 \) into the margin of error formula:
\( E = 1.333\times\frac{7.1}{\sqrt{18}} \)
First, calculate \( \sqrt{18}\approx4.2426 \)
Then, \( \frac{7.1}{4.2426}\approx1.6735 \)
Then, \( E=1.333\times1.6735\approx2.23 \) (rounded to two decimal places)

Step4: Calculate the confidence interval

The confidence interval is \( \bar{x}\pm E \), so \( 34.7\pm2.23 \)
Lower bound: \( 34.7 - 2.23=32.47 \)
Upper bound: \( 34.7+2.23 = 36.93 \)

Answer:

Margin of error: \( \approx2.23 \)
Confidence interval: \( (32.5, 36.9) \) (rounded to one decimal place as needed)