Question

wildlife biologists inspect 146 deer taken by hunters and find 35 of them carrying ticks that test positive for lyme disease. a) create a 90% confidence interval for the percentage of deer that may carry such ticks. (18.2%, 29.8%) (round to one decimal place as needed.) b) if the scientists want to cut the margin of error in half, how many deer must they inspect?

Explanation:

Step1: Calculate sample proportion

$\hat{p} = \frac{35}{146} \approx 0.2397$

Step2: Find z-score for 90% CI

For 90% confidence, $z_{\alpha/2} = 1.645$

Step3: Compute margin of error (a)

$E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 1.645\sqrt{\frac{0.2397(1-0.2397)}{146}} \approx 0.0585$

Step4: Calculate CI bounds (a)

Lower bound: $\hat{p} - E = 0.2397 - 0.0585 = 0.1812 \approx 18.1\%$
Upper bound: $\hat{p} + E = 0.2397 + 0.0585 = 0.2982 \approx 29.8\%$

Step5: Relate sample size to E (b)

To halve E, $n$ scales by $4$. New $n = 146 \times 4 = 584$

Answer:

a) (18.1 %, 29.8 %)
b) 584