Question

in a random sample of 23 people, the mean commute time to work was 31.8 minutes and the standard deviation was 7.2 minutes. assume the population is normally distributed and use a t - distribution to construct a 95% 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 (28.7, 34.9).
(round to one decimal place as needed.)

the margin of error of μ is
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for margin of error (E)

The margin of error is related to the confidence interval. The confidence interval for the population mean \(\mu\) is given by \(\bar{x} \pm E\), where \(\bar{x}\) is the sample mean. Also, the confidence interval can be written as \((\text{Lower bound}, \text{Upper bound})\). So, we can find \(E\) using the formula \(E=\frac{\text{Upper bound}-\text{Lower bound}}{2}\) or \(E = \bar{x}-\text{Lower bound}\) or \(E=\text{Upper bound}-\bar{x}\).

We know that \(\bar{x} = 31.8\) minutes, the lower bound is \(28.7\) minutes, and the upper bound is \(34.9\) minutes.

Step2: Calculate the margin of error

Using the formula \(E=\bar{x}-\text{Lower bound}\):
\(E = 31.8 - 28.7 = 3.1\)
Or using the formula \(E=\frac{\text{Upper bound}-\text{Lower bound}}{2}\):
\(E=\frac{34.9 - 28.7}{2}=\frac{6.2}{2}=3.1\)

Answer:

\(3.1\)