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If n=540 and

(p-hat) = 0.64, construct a 90% confidence interval.

Give your answers to three decimals.

__ < p < __

1 Answer

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Answer:

0.606 < p < 0.674

Explanation:


\displaystyle CI=\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\\\CI_(90\%)=0.64\pm1.645\sqrt{(0.64(1-0.64))/(540)}\\\\CI_(90\%)\approx0.64\pm0.034\\\\CI_(90\%)=\{0.606,0.674\}

Thus, we are 90% confident that the true population proportion is between 0.606 and 0.674

User Mrjasmin
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