Question #263683

Determine the sample size necessary to estimate a population proportion to within 0.03 with 90%

confidence assuming you have no knowledge of the approximate value of the population proportion.

(b)    Using the sample size obtained in (a) and pˆ = 0.75

Estimate the population with 90% confidence


Expert's answer

(a):

Margin of error (E) = 0.03

Z value at 90% confidence interval = 1.645

E=Z×p×(1−p)nE=Z\times\sqrt{\frac{p\times(1-p)}{n}}


0.03=1.645×0.5×(1−0.5)n0.03=1.645\times\sqrt{\frac{0.5\times(1-0.5)}{n}}


Square both sides;


0.0009=2.71×0.5×(1−0.5)n0.0009=2.71\times\frac{0.5\times(1-0.5)}{n}


n=2.71×0.250.0009=751.67n= \frac{2.71\times0.25}{0.0009}=751.67

Sample size = 752


(b):

pˆ = 0.75

sample size = 752

p±Z×p×(1−p)np\pm Z\times\sqrt{\frac{p\times(1-p)}{n}}


0.75±1.645×0.75×(1−0.75)7520.75\pm 1.645\times\sqrt{\frac{0.75\times(1-0.75)}{752}}


0.75±0.0260.75\pm0.026


0.724 to 0.776




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