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In a recent poll of 200 households, it was found that 152 had at least one computer and one television. A 95% confidence interval to estimate the...


In a recent poll of 200​ households, it was found that 152 had at least one computer and one television. A​ 95% confidence interval to estimate the population proportion was calculated to be 0.701 to 0.819. What sample size would be needed to change the margin of error to​ 0.030?

A.

534

B.

779

C.

1068

D.

390

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2 comments
  • Correction, n = 779
    • chinnaistats
    • May 06, 2018 at 5:49pm
  • (1.96 / 0.030)^2 * 0.76*0.24 = 779
    • chinnaistats
    • May 06, 2018 at 5:57pm

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