Question #189572

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.

1
Expert's answer
2021-05-07T14:16:30-0400

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.

p^=0.75n=280α=10.99=0.01\hat{p} = 0.75 \\ n = 280 \\ α = 1 - 0.99 = 0.01

Crotocal z-value

zα/2=z0.01/2=z0.005=2.58z_{α/2} = z_{0.01/2} = z_{0.005} = 2.58

Standard error of p^:\hat{p}:

SE=p^×(1p^)n=0.75×0.25280=0.025877E=zα/2×SE=2.58×0.025877=0.066764SE = \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(p^(1p^)n)0.5E = Z( \frac{\hat{p}(1- \hat{p})}{n} )^{0.5}

z = 1.96 for 95 % confidence

p^=Xn=6603300=0.20n=3300E=1.96(0.2×0.83300)0.5=0.0136\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


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