Question

overtime hours worked a random sample of 14 registered nurses in a large hospital showed that they worked on average 44.6 hours per week. the standard deviation of the sample was 2.6. estimate the mean of the population with 99% confidence. assume the variable is normally distributed. round intermediate answers to at least three decimal places. round your final answers to one decimal place.

Explanation:

Step1: Determine the degrees of freedom and t - value

The degrees of freedom \(df=n - 1\), where \(n = 14\), so \(df=14 - 1=13\).
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 \(t_{\frac{\alpha}{2},df}=t_{0.005,13}=3.012\) (rounded to three decimal places).

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 \(s = 2.6\) (sample standard deviation) and \(n = 14\) (sample size).
Substitute the values: \(E=3.012\times\frac{2.6}{\sqrt{14}}\).
First, calculate \(\sqrt{14}\approx3.742\).
Then \(\frac{2.6}{3.742}\approx0.695\).
So \(E = 3.012\times0.695\approx2.1\) (rounded to one decimal place).

Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), where \(\bar{x}=44.6\) (sample mean).
\(\bar{x}-E=44.6 - 2.1=42.5\).
\(\bar{x}+E=44.6+2.1 = 46.7\).

Answer:

\(42.5<\mu<46.7\)