in a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. the…

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...

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...

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 z - value for 68% confidence interval

For a 68% confidence interval, the z - value (z) (from the standard normal distribution) is (z = 1) (since approximately 68% of the data lies within (z=\pm1) in a normal distribution).

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

The formula for the confidence interval for a proportion is (\hat{p}\pm z\times SE)

  • Lower bound: (p - z\times SE=0.301-1\times0.0503 = 0.2507)
  • Upper bound: (p + z\times SE=0.301 + 1\times0.0503=0.3513)

Answer:

(0.2507) to (0.3513)