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. (□% , □%) (round to one decimal place as needed.)

Explanation:

Step1: Calculate sample proportion

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

Step2: Find critical z-value

For 90% confidence, $z^* = 1.645$

Step3: Compute standard error

$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.2397(1-0.2397)}{146}} \approx \sqrt{\frac{0.1822}{146}} \approx 0.0353$

Step4: Calculate margin of error

$ME = z^* \times SE = 1.645 \times 0.0353 \approx 0.0581$

Step5: Find confidence interval

Lower bound: $0.2397 - 0.0581 = 0.1816 \approx 18.2\%$
Upper bound: $0.2397 + 0.0581 = 0.2978 \approx 29.8\%$

Answer:

(18.2%, 29.8%)