Question

monthly income of d.c. workers a survey of 15 randomly selected employed people in washington, d.c. found that the mean monthly income was $6250. the standard deviation for the sample is $694. find the 99% confidence interval for the mean monthly income of the workers in washington, d.c. round intermediate calculations to three decimal places, and final answers to the nearest whole number. $ < μ < $

Explanation:

Step1: Determine the degrees of freedom and t - value

The degrees of freedom \(df=n - 1\), where \(n = 15\), so \(df=15 - 1=14\).
For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99 = 0.01\), and \(\frac{\alpha}{2}=0.005\).
Using the t - distribution table or a calculator, the t - value \(t_{\frac{\alpha}{2},df}=t_{0.005,14}=2.977\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a confidence interval for the population mean when the population standard deviation \(\sigma\) is unknown is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 694\), \(n = 15\), and \(t_{\frac{\alpha}{2}}=2.977\).

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Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=6250\).
\(\bar{x}-E=6250 - 533.5=5716.5\approx5717\)
\(\bar{x}+E=6250+533.5 = 6783.5\approx6784\)

Answer:

\(5717<\mu<6784\)