## Related questions with answers

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

Solution

VerifiedGiven: Confidence interval for $p$

$(0.57,0.63)$

$c=50\%=0.5$

$\dfrac{k}{n}$ 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:

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

For confidence level $1-\alpha=0.5$, determine $z_{\alpha/2}=z_{0.25}$ 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):

$z_{\alpha/2}=0.67$

The margin of error is then:

$E=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=0.67\times \sqrt{\dfrac{0.60(1-0.60)}{n}}=0.67\times \sqrt{\dfrac{0.24}{n}}$

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

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

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