## Related questions with answers

Wildlife biologists inspect 153 deer taken by hunters and find 32 of them carrying ticks that test positive for Lyme disease. If the scientists want to cut the margin of error in half, how many deer must they inspect?

Solution

VerifiedGiven:

$x=32$

$n=153$

$c=90\%=0.90$

The sample proportion is the number of successes divided by the sample size:

$\hat{p}=\dfrac{x}{n}=\dfrac{32}{153}\approx 0.2092$

For confidence level $1-\alpha=0.90$, determine $z_{\alpha/2}=z_{0.05}$ using table Z in appendix F (look up 0.05 in the table, the z-score is then the found z-score with opposite sign):

$z_{\alpha/2}=1.645$

The margin of error is then:

$E=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=1.645\cdot \sqrt{\dfrac{0.2092(1-0.2092)}{153}}\approx 0.0541$

We want to half the margin of error:

$E_N=\dfrac{0.0541}{2}=0.02705$

The sample size is then (round up!):

$n=\dfrac{[z_{\alpha/2}]^2\hat{p}(1-\hat{p})}{E^2}=\dfrac{1.645^2 (0.2092)(1-0.2092)}{0.02705^2}\approx 612$

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