Question

vacation days during a recent year, working americans took on average 17.4 vacation days. this survey was based on a random sample of 19 workers. the standard deviation of the sample was 2.3 days. find the 95% confidence interval for the mean number of vacation days workers took per year. assume the population is normally distributed. use the t distribution table. round intermediate answers to at least three decimal places. round your final answers to the nearest whole number.

Explanation:

Step1: Determine the degrees of freedom

Degrees of freedom \(df=n - 1\), where \(n = 19\). So \(df=19-1 = 18\)

Step2: Find the critical value \(t_{\alpha/2}\)

For a \(95\%\) confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). Using the \(t\)-distribution table with \(df = 18\), \(t_{0.025}=2.101\)

Step3: Calculate the margin of error \(E\)

The formula for the margin of error for a \(t\)-distribution is \(E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}\), where \(s = 2.3\) and \(n = 19\)

$$
LATEXBLOCK0
$$

Step4: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=17.4\)

$$
LATEXBLOCK1
$$

Rounding to the nearest whole number, we get \(16<\mu<19\)

Answer:

$$16 < \mu < 19$$