Question

in a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. the 68% confidence interval for the proportion of persons who work less than 40 hours per week is type your answer... to type your answer...

Explanation:

Step1: Calculate the standard error

The formula for the standard error of a proportion is \(SE = \sqrt{\frac{p(1 - p)}{n}}\), where \(p = 0.301\) and \(n=83\).

$$SE=\sqrt{\frac{0.301\times(1 - 0.301)}{83}}=\sqrt{\frac{0.301\times0.699}{83}}\approx\sqrt{\frac{0.2104}{83}}\approx\sqrt{0.002535}\approx0.0503$$

Step2: Find the confidence interval

For a 68% confidence interval, the z - value (from the standard normal distribution) is \(z = 1\) (since about 68% of the data lies within \(z=\pm1\) in a normal distribution).
The confidence interval is given by \(\hat{p}\pm z\times SE\)
The lower limit is \(0.301-1\times0.0503 = 0.2507\)
The upper limit is \(0.301 + 1\times0.0503=0.3513\)

Answer:

\(0.251\) to \(0.351\)