Question

1. In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 68 percent confidence interval for the proportion of persons who work less than 40 hours per week.

Answer 1

To calculate the 68 percent confidence interval for the proportion of persons working less than 40 hours per week from a sample of 83 respondents with a sample proportion of 30.1 percent, we use the formula for the confidence interval for a proportion. The resulting interval is approximately 25.06% to 35.14%.

Step-by-step explanation:

To calculate a 68 percent confidence interval for the proportion of persons who work less than 40 hours per week from a sample of 83 respondents, where 30.1 percent work less than 40 hours, we use the formula for a confidence interval for a proportion:

In this formula:

• p is the sample proportion (0.301 in this case).
• z* is the z-value corresponding to the desired confidence level (for 68 percent confidence, use the z-value corresponding to one standard deviation from the mean in a standard normal distribution, which is approximately 1).
• n is the sample size (83).

Plugging the values into the formula we get:

0.301±1*sqrt((0.301(0.699)/83))

Calculating the square root part, we have:

0.301±1*sqrt((0.301*0.699)/83)
= 0.301±1*sqrt(0.210699/83)
= 0.301±1*sqrt(0.002539)
= 0.301±1*0.05039
= 0.301±0.05039

The confidence interval is thus:

0.301-0.05039 to 0.301+0.05039
= 0.25061 to 0.35139

Hence, with a 68 percent confidence level, we can say that the true proportion of the population that works less than 40 hours per week is estimated to be between 25.06% and 35.14%.

Answer 2

A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)

Step-by-step explanation:

We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is
`\hat{p} = 0.301` because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of
`\hat{p}` as
`\sqrt{\hat{p}(1-\hat{p})/n}=√(0.301(1-0.301)/83) = 0.0503`. A
`100(1-\alpha)%` confidence interval is given by
`\hat{p}\pm z_(\alpha/2)\sqrt{\hat{p}(1-\hat{p})/n`, then, a 68% confidence interval is
`0.301\pm z_(0.32/2)0.0503`, i.e.,
`0.301\pm (0.9944)(0.0503)`, i.e., (0.251, 0.351).
`z_(0.16) = 0.9944` is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.