2. Wildlife biologists inspect 153 deer taken by hunters and find 32 of them carrying Lyme disease ticks. d) If the scientists want to cut the margin of error, E in half, calculate the number of deer that must be inspected? Consider the same 98% C.I. Show all mathematics.

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**Wildlife Biology Study on Lyme Disease Ticks**

In a recent study conducted by wildlife biologists, 153 deer taken by hunters were inspected, and 32 of these deer were found to be carrying Lyme disease ticks.

**Cutting the Margin of Error in Half**

**Problem Statement:**
If the scientists want to cut the margin of error (\(E\)) in half, calculate the number of deer that must be inspected. Consider the same 98% Confidence Interval (C.I.). Show all mathematics.

---

For educational purposes, let us go through the steps to determine the required sample size when the margin of error is halved, keeping the confidence level constant at 98%.

### Step-by-Step Solution:

**Step 1: Initial conditions and formula for margin of error**
The margin of error (\(E\)) for a proportion can be defined as:
\[ E = z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:
- \( z \) is the z-value for the desired confidence level
- \( \hat{p} \) is the sample proportion (in this case, \(\hat{p} = \frac{32}{153}\))
- \( n \) is the sample size

**Step 2: Finding the sample proportion**
\[ \hat{p} = \frac{32}{153} \approx 0.209 \]

**Step 3: Determining the z-value for a 98% Confidence Interval**
From standard z-tables, the z-value for a 98% confidence level is approximately 2.33.

**Step 4: Writing the initial margin of error equation (E)**
\[ E = 2.33 \times \sqrt{\frac{0.209 \times (1 - 0.209)}{153}} \]

Calculating the value inside the square root:
\[ 0.209 \times (1 - 0.209) = 0.209 \times 0.791 \approx 0.165 \]

Then, 
\[ E = 2.33 \times \sqrt{\frac{0.165}{153}} \approx 2.33 \times \sqrt{0.00108} \approx 2.33 \times 0.0328 \approx 0.076 \]

**Step 5: Halving the margin of error (\(
Transcribed Image Text:**Wildlife Biology Study on Lyme Disease Ticks** In a recent study conducted by wildlife biologists, 153 deer taken by hunters were inspected, and 32 of these deer were found to be carrying Lyme disease ticks. **Cutting the Margin of Error in Half** **Problem Statement:** If the scientists want to cut the margin of error (\(E\)) in half, calculate the number of deer that must be inspected. Consider the same 98% Confidence Interval (C.I.). Show all mathematics. --- For educational purposes, let us go through the steps to determine the required sample size when the margin of error is halved, keeping the confidence level constant at 98%. ### Step-by-Step Solution: **Step 1: Initial conditions and formula for margin of error** The margin of error (\(E\)) for a proportion can be defined as: \[ E = z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] Where: - \( z \) is the z-value for the desired confidence level - \( \hat{p} \) is the sample proportion (in this case, \(\hat{p} = \frac{32}{153}\)) - \( n \) is the sample size **Step 2: Finding the sample proportion** \[ \hat{p} = \frac{32}{153} \approx 0.209 \] **Step 3: Determining the z-value for a 98% Confidence Interval** From standard z-tables, the z-value for a 98% confidence level is approximately 2.33. **Step 4: Writing the initial margin of error equation (E)** \[ E = 2.33 \times \sqrt{\frac{0.209 \times (1 - 0.209)}{153}} \] Calculating the value inside the square root: \[ 0.209 \times (1 - 0.209) = 0.209 \times 0.791 \approx 0.165 \] Then, \[ E = 2.33 \times \sqrt{\frac{0.165}{153}} \approx 2.33 \times \sqrt{0.00108} \approx 2.33 \times 0.0328 \approx 0.076 \] **Step 5: Halving the margin of error (\(
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