there is no prior information about the proportion of americans who support free trade in 2018. if we want to estimate a 97.5% confidence interval for the true proportion of americans who support free trade in 2018 with a 0.16 margin of error, how many randomly selected americans must be surveyed? answer: (round up your answer to nearest whole number)

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SidHussain SidHussain · Answer 1 · 2022-10-14T17:34:45+02:00

The number of randomly selected Americans that must be surveyed is equal to 38 if the confidence interval is 97.5% with a 0.16 margin of error.

First, we calculate the critical value as follows;

a = (1 - 0.975) = 0.025 = 0.025 / 2 = 0.0125 = (1 - 0.0125) = 0.9875

Z - Critical value = NORM.S.INV (0.9875) = 1.96

As margin of error(ME) = 0.16

Therefore;

n = pq(Z²) / ME²

n = 0.5 × 0.5 (1.96)² / (0.16)²

n = 0.25 × 3.8416 / 0.0256

n = 0.9604 / 0.0256

n = 37.515

Rounding it to the nearest whole number;

n = 37.515 = 38

Hence 38 randomly selected Americans must be surveyed if the estimated confidence interval is 97.5% and the margin of error is 0.16.

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