suppose the consumer research center researchers wanted to be 95% confident that teir estimate is within 2% of the true population proportion. how large a sample should they choose for their survey? the preliminary estimate was 29%
Solution :
Given that,
= 29%=0.29
1 - = 1 - 0. 29= 0.71
margin of error = E = 2% = 0.02
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Sample size = n = (Z/2 / E)2 * * (1 - )
= (1.96 / 0.02)2 * 0.29* 0.71
= 1977.4636
Sample size =1978
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