Answer to Question #189572 in Statistics and Probability for TANYA

Question

In a survey of 280 adults over 50, 75% said they were taking vitamin supplements. Find the margin of error for this survey if we want a 99% confidence in our estimate of the percent of adults over 50 who take vitamin supplements.

In a survey of 3300 T.V. viewers, 20% said they watch the network news programs. Find the margin of error for this survey if we want a 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs.

Answer 1

In a survey of 280 adults over 50, 75% said they were taking vitamin supplements. Find the margin of error for this survey if we want a 99% confidence in our estimate of the percent of adults over 50 who take vitamin supplements.

$\hat{p} = 0.75 \\ n = 280 \\ α = 1 - 0.99 = 0.01$

Crotocal z-value

$z_{α/2} = z_{0.01/2} = z_{0.005} = 2.58$

Standard error of $\hat{p}:$

$SE = \sqrt{\frac{\hat{p} \times (1- \hat{p})}{n}} \\ = \sqrt{\frac{0.75 \times 0.25}{280}} \\ = 0.025877 \\ E = z_{α/2} \times SE \\ = 2.58 \times 0.025877 \\ = 0.066764$

99 % Confidence interval:

0.75 ± 0.066764

(0.683, 0.817)

In a survey of 3300 T.V. viewers, 20% said they watch the network news programs. Find the margin of error for this survey if we want a 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs.

The equation for the margin of error is:

$E = Z( \frac{\hat{p}(1- \hat{p})}{n} )^{0.5}$

z = 1.96 for 95 % confidence

$\hat{p} = \frac{X}{n} = \frac{660}{3300} = 0.20 \\ n = 3300 \\ E = 1.96( \frac{0.2 \times 0.8}{3300} )^{0.5} = 0.0136$