Wildlife biologists inspect 156 deer taken by hunters and find 31 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. ( $$ 14.6 \%, 25.1 \% $$ ) (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?

Question

UpStudy AI · Accepted Answer

To create a $$ 90\% $$ confidence interval for the percentage of deer that may carry ticks testing positive for Lyme disease, we can follow these steps:

### Part a: Confidence Interval Calculation

1. **Calculate the sample proportion ($$ \hat{p} $$)**:
   $$
   \hat{p} = \frac{x}{n} = \frac{31}{156} \approx 0.1987
   $$

2. **Calculate the standard error (SE)**:
   $$
   SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.1987(1 - 0.1987)}{156}} \approx \sqrt{\frac{0.1987 \times 0.8013}{156}} \approx \sqrt{\frac{0.1595}{156}} \approx \sqrt{0.001021} \approx 0.0319
   $$

3. **Find the z-score for a $$ 90\% $$ confidence level**:
   The z-score for a $$ 90\% $$ confidence level is approximately $$ 1.645 $$.

4. **Calculate the margin of error (ME)**:
   $$
   ME = z \times SE = 1.645 \times 0.0319 \approx 0.0525
   $$

5. **Calculate the confidence interval**:
   $$
   \text{Lower limit} = \hat{p} - ME \approx 0.1987 - 0.0525 \approx 0.1462
   $$
   $$
   \text{Upper limit} = \hat{p} + ME \approx 0.1987 + 0.0525 \approx 0.2512
   $$

6. **Convert to percentage**:
   $$
   \text{Lower limit} \approx 14.6\%
   $$
   $$
   \text{Upper limit} \approx 25.1\%
   $$

Thus, the $$ 90\% $$ confidence interval for the percentage of deer that may carry ticks testing positive for Lyme disease is approximately $$ (14.6\%, 25.1\%) $$.

### Part b: Sample Size Calculation for Half Margin of Error

To cut the margin of error in half, we need to determine the new sample size ($$ n $$).

1. **Current margin of error (ME)**:
   $$
   ME = 0.0525
   $$
   To cut this in half:
   $$
   \text{New ME} = \frac{0.0525}{2} = 0.02625
   $$

2. **Use the formula for margin of error**:
   $$
   ME = z \times SE = z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
   $$
   Rearranging for $$ n $$:
   $$
   n = \frac{z^2 \hat{p}(1 - \hat{p})}{ME^2}
   $$

3. **Substituting the values**:
   $$
   n = \frac{(1.645)^2 \times 0.1987 \times (1 - 0.1987)}{(0.02625)^2}
   $$
   $$
   n = \frac{2.7060 \times 0.1987 \times 0.8013}{0.000689}
   $$
   $$
   n = \frac{2.7060 \times 0.1595}{0.000689} \approx \frac{0.4313}{0.000689} \approx 626.5
   $$

Since we cannot inspect a fraction of a deer, we round up to the nearest whole number:
$$
n \approx 627
$$

Thus, the scientists must inspect **at least 627 deer** to cut the margin of error in half.
a) The $$ 90\% $$ confidence interval is $$ (14.6\%, 25.1\%) $$. b) To cut the margin of error in half, scientists need to inspect at least 627 deer.

Real Tutor Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Statistics Questions