Question

translate the statement into a confidence interval. approximate the level of confidence.
in a survey of 1015 parents of children ages 8 - 14, 63% say they are willing to get a second or part - time job to pay for their childrens college education and 36% say they lose sleep worrying about college costs. the surveys margin of error is ±3%.
use the results from the survey about getting a second or part - time job.
the confidence interval for the proportion is (0.600, 0.660).
(round to three decimal places as needed.)
use the results from the survey about getting a second or part - time job.
the interval is a 95.2% confidence interval.
(round to one decimal place as needed.)
use the results from the survey about losing sleep.
the confidence interval for the proportion is (0.330, 0.390).
(round to three decimal places as needed.)
use the results from the survey about losing sleep.
the interval is a □% confidence interval.
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the margin of error formula for proportion

The formula for the margin of error \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For the first case (63% with \(n = 1015\)), \(E=0.03\), \(p = 0.63\). But we can also use the fact that for a proportion confidence interval \(\hat{p}-EWe know that for a proportion, the margin of error \(E=z\sqrt{\frac{p(1 - p)}{n}}\). Substituting \(n = 1015\), \(p = 0.36\), \(E=0.03\)

$$z=\frac{E}{\sqrt{\frac{p(1 - p)}{n}}}$$

$$z=\frac{0.03}{\sqrt{\frac{0.36\times(1 - 0.36)}{1015}}}$$

$$z=\frac{0.03}{\sqrt{\frac{0.36\times0.64}{1015}}}$$

$$z=\frac{0.03}{\sqrt{\frac{0.2304}{1015}}}$$

$$z=\frac{0.03}{\sqrt{0.000227}}$$

$$z=\frac{0.03}{0.0151}$$

$$z\approx1.99$$

Step2: Find the confidence level

Looking up the \(z\) - value in the standard normal distribution table. The \(z\) - value of \(z = 1.99\) corresponds to a confidence level. The confidence level \(C\) is related to the \(z\) - value by the formula \(C = 2\Phi(z)-1\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.
From the standard normal table, \(\Phi(1.99)=0.9767\)

$$C=2\times0.9767 - 1$$

$$C = 0.9534\approx95.3\%$$

Answer:

\(95.3\%\)