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A research center survey of 2,351 adults found that 1,899 had bought something online. Of these online shoppers, 1,203 are we

a. Construct a​ 95% confidence interval estimate of the population proportion of adults who had bought something online

b. Construct a​ 95% confidence interval estimate of the population proportion of online shoppers who are weekly online shoppers.

math   Statistics-And-Probability   
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Answer #1

Part a)

p̂ = X / n = 1899/2351 = 0.8077
p̂ ± Z(α/2) √( (p * q) / n)
0.8077 ± Z(0.05/2) √( (0.8077 * 0.1923) / 2351)
Critical value Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.8077 - Z(0.05) √( (0.8077 * 0.1923) / 2351) = 0.7918
upper Limit = 0.8077 + Z(0.05) √( (0.8077 * 0.1923) / 2351) = 0.8237
95% Confidence interval is ( 0.7918 , 0.8237 )
( 0.7918 < P < 0.8237 )

Part b)

p̂ = X / n = 1203/1899 = 0.6335
p̂ ± Z(α/2) √( (p * q) / n)
0.6335 ± Z(0.05/2) √( (0.6335 * 0.3665) / 1899)
Critical value Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.6335 - Z(0.05) √( (0.6335 * 0.3665) / 1899) = 0.6118
upper Limit = 0.6335 + Z(0.05) √( (0.6335 * 0.3665) / 1899) = 0.6552
95% Confidence interval is ( 0.6118 , 0.6552 )
( 0.6118 < P < 0.6552 )

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