Amphibian populations are in decline worldwide due to the increasing prevalence of ranavirus infections. Herpetologists (those that study amphibians and reptiles) are especially concerned with how ranavirus infections may be affecting Ozark and eastern hellbender populations.

Suppose both populations of hellbenders are known to have ranavirus infections at a rate of 25%. Suppose a random sample of size 232 is taken of the eastern hellbender population and a random sample of size 348 is taken of the Ozark hellbender population. Every hellbender in each sample is tested for ranavirus, and the sample proportions for each of the two samples are calculated.

Which of the two sample proportions is more likely to exceed 22% ?


Question 3Answer

a.
They are equally likely to exceed 22%. 0

b.
The sample proportion of ranavirus infections for the eastern hellbenders. 0

c.
There is not enough information present to answer this question. 0

d.
The sample proportion of ranavirus infections for the Ozark hellbenders. 0

To answer this question, we need to analyze which sample proportion (for eastern hellbenders or Ozark hellbenders) is more likely to exceed 22%, given that the true infection rate is 25% for both populations.

### Step 1: Calculate the standard error for each sample proportion
The standard error (SE) for the proportion is calculated using the formula:

\[
SE = \sqrt{\frac{p(1 - p)}{n}}
\]

where:
- \(p\) is the true proportion of infections (25% or 0.25 in this case),
- \(n\) is the sample size.

#### Eastern Hellbenders:
- \(p = 0.25\),
- \(n = 232\).

The standard error for the eastern hellbender sample proportion:

\[
SE_{\text{east}} = \sqrt{\frac{0.25(1 - 0.25)}{232}} = \sqrt{\frac{0.25 \times 0.75}{232}} \approx \sqrt{\frac{0.1875}{232}} \approx \sqrt{0.000808} \approx 0.0284
\]

#### Ozark Hellbenders:
- \(p = 0.25\),
- \(n = 348\).

The standard error for the Ozark hellbender sample proportion:

\[
SE_{\text{ozark}} = \sqrt{\frac{0.25(1 - 0.25)}{348}} = \sqrt{\frac{0.1875}{348}} \approx \sqrt{0.000539} \approx 0.0232
\]

### Step 2: Compare the probabilities of exceeding 22%

Next, we calculate the **z-score** for 22% in each case. The z-score formula is:

\[
z = \frac{\hat{p} - p}{SE}
\]

where:
- \(\hat{p}\) is the proportion we are testing against (22% or 0.22),
- \(p\) is the true proportion (25% or 0.25),
- \(SE\) is the standard error.

#### Eastern Hellbenders:
For the eastern hellbenders, the z-score is:

\[
z_{\text{east}} = \frac{0.22 - 0.25}{0.0284} = \frac{-0.03}{0.0284} \approx -1.06
\]

#### Ozark Hellbenders:
For the Ozark hellbenders, the z-score is:

\[
z_{\text{ozark}} = \frac{0.22 - 0.25}{0.0232} = \frac{-0.03}{0.0232} \approx -1.29
\]

### Step 3: Interpret the z-scores
The z-scores tell us how many standard deviations below the mean (0.25) the value 0.22 is for each population. A lower z-score indicates that it's less likely to fall below the value of 22%.

- Eastern hellbenders: \(z_{\text{east}} \approx -1.06\)
- Ozark hellbenders: \(z_{\text{ozark}} \approx -1.29\)

Since the z-score for the **eastern hellbenders** is closer to 0, it means that the sample proportion for the **eastern hellbenders** is more likely to exceed 22% compared to the Ozark hellbenders.

### Conclusion:
The correct answer is:

**b. The sample proportion of ranavirus infections for the eastern hellbenders.**

---

Would you like more details or have any other questions?

Here are 5 related questions:
1. How can z-scores be used to compare probabilities in different populations?
2. What is the significance of the standard error in estimating proportions?
3. How does sample size affect the likelihood of exceeding a certain proportion?
4. How is the central limit theorem relevant to this type of problem?
5. What assumptions do we make about the distribution of sample proportions?

**Tip:** Larger sample sizes reduce the standard error, making estimates of population proportions more precise.

Amphibian populations are in decline worldwide due to the increasing prevalence of ranavirus infections. Herpetologists (those that study amphibians and reptiles) are especially concerned with how ranavirus infections may be affecting Ozark and eastern hellbender populations. Suppose both populations of hellbenders are known to have ranavirus infections at a rate of 25%. Suppose a random sample of size 232 is taken of the eastern hellbender population and a random sample of size 348 is taken of the Ozark hellbender population. Every hellbender in each sample is tested for ranavirus, and the sample proportions for each of the two samples are calculated. Which of the two sample proportions is more likely to exceed 22%?

Explore a statistical comparison of ranavirus infection rates in eastern and Ozark hellbender populations. Using sample proportions, standard error, and z-scores, this problem examines which population is more likely to have an infection rate exceeding 22%. Suitable for advanced high school or college students, this detailed analysis includes step-by-step solutions and key statistical concepts.

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