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Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 221
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Answer #1

We would be looking at the first question here as:

From the standard normal tables, we have:
P(Z < 1.282) = 0.9

Therefore, due to symmetry, we get here:
P( -1.282 < Z < 1.282) = 0.8

The sample proportion here is computed as:
p = x/n = 75/221 = 0.3394

The 80% confidence interval for proportion here is computed as:

p \pm z^*\sqrt{\frac{p(1-p)}{n}}

0.3394 \pm 1.282*\sqrt{\frac{0.3394(1-0.3394)}{221}}

0.3394 \pm 0.0408

(0.299, 0.380)

This is the required 80% confidence interval here.

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