Wildlife biologists inspect 156 deer taken by hunters and find 31 of them carrying ticks that test positive for Lyme disease.
a) Create a confidence interval for the percentage of deer that may carry such ticks.
( )
(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?

Ask by Norris Schwartz. in the United States

Nov 06,2024

Question

Wildlife biologists inspect 156 deer taken by hunters and find 31 of them carrying ticks that test positive for Lyme disease.
a) Create a confidence interval for the percentage of deer that may carry such ticks.
( )
(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?

Ask by Norris Schwartz. in the United States

Nov 06,2024

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.

Upstudy AI Solution

Answer

Solution

Part a: Confidence Interval Calculation

Part b: Sample Size Calculation for Half Margin of Error

Bonus Knowledge

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