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 $n=83$.
$$SE=\sqrt{\frac{0.301\times(1 - 0.301)}{83}}=\sqrt{\frac{0.301\times0.699}{83}}\approx\sqrt{\frac{0.210399}{83}}\approx\sqrt{0.002535}\approx0.0503$$

Step2: Find the margin of error

For a 68% confidence interval, the z - score $z = 1$ (from the standard normal distribution, since about 68% of the data lies within 1 standard deviation of the mean). The margin of error $ME=z\times SE$.
So, $ME = 1\times0.0503=0.0503$

Step3: Calculate the lower and upper bounds of the confidence interval

The lower bound is $p - ME$ and the upper bound is $p+ME$.
Lower bound: $0.301-0.0503 = 0.2507$
Upper bound: $0.301 + 0.0503=0.3513$

Answer:

$0.251$ to $0.351$