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If (0.57, 0.63) is a 50% confidence interval for p, what does kn\frac{k}{n} equal, and how many observations were taken?

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Given: Confidence interval for pp

(0.57,0.63)(0.57,0.63)

c=50%=0.5c=50\%=0.5

kn\dfrac{k}{n} represents the sample proportion, which lies exactly in the middle of the confidence interval and thus is equal to the average of the boundaries of the confidence interval:

kn=0.57+0.632=0.60\dfrac{k}{n}=\dfrac{0.57+0.63}{2}=0.60

For confidence level 1α=0.51-\alpha=0.5, determine zα/2=z0.25z_{\alpha/2}=z_{0.25} using using the normal probability table in the appendix (look up 0.25 in the table, the z-score is then the found z-score with opposite sign):

zα/2=0.67z_{\alpha/2}=0.67

The margin of error is then:

E=zα/2p^(1p^)n=0.67×0.60(10.60)n=0.67×0.24nE=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=0.67\times \sqrt{\dfrac{0.60(1-0.60)}{n}}=0.67\times \sqrt{\dfrac{0.24}{n}}

The margin of error is also half the width of the confidence interval:

E=0.630.572=0.03E=\dfrac{0.63-0.57}{2}=0.03

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