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 ( ).
(round to one decimal place as needed.)

Explanation:

Step1: Identify t - value

We have a sample size \(n = 23\), so the degrees of freedom \(df=n - 1=23 - 1 = 22\). For a 95% confidence interval, the significance level \(\alpha=1 - 0.95 = 0.05\), and \(\alpha/2=0.025\). Looking up the t - value in the t - distribution table for \(df = 22\) and \(\alpha/2=0.025\), we get \(t_{\alpha/2}=t_{0.025,22}\approx2.074\).

Step2: Calculate margin of error

The formula for the margin of error \(E\) when using the t - distribution is \(E = t_{\alpha/2}\times\frac{s}{\sqrt{n}}\), where \(s\) is the sample standard deviation, \(n\) is the sample size. Here, \(s = 7.2\), \(n = 23\), and \(t_{\alpha/2}\approx2.074\).
First, calculate \(\sqrt{n}=\sqrt{23}\approx4.796\). Then, \(\frac{s}{\sqrt{n}}=\frac{7.2}{4.796}\approx1.501\).
Now, \(E=2.074\times1.501\approx3.1\) (rounded to one decimal place).

Step3: Calculate confidence interval

The formula for the confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}\) is the sample mean. Here, \(\bar{x}=31.8\) and \(E\approx3.1\).
Lower bound: \(31.8-3.1 = 28.7\)
Upper bound: \(31.8 + 3.1=34.9\)

Answer:

The confidence interval for the population mean \(\mu\) is \(28.7<\mu<34.9\) (or in the boxed form as \((28.7, 34.9)\)). The margin of error is approximately \(3.1\) minutes. This means we are 95% confident that the true population mean commute time lies between \(28.7\) minutes and \(34.9\) minutes.