Mathematics, 22.02.2021 19:20 ashiteru123

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 118 is taken of the eastern hellbender population and a random sample of size 177 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.

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

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Answer from: inneedofhelp6237

The sample proportion of size 177 is more likely to exceed 22%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose both populations of hellbenders are known to have ranavirus infections at a rate of 25%.

This means that $\mu = p = 0.25$

Sample of size 118:

$n = 118, so [tex]s = \sqrt{\frac{0.25*0.75}{118}} = 0.0399$

Probability of sample proportion above 22%.

This is 1 subtracted by the pvalue of Z when X = 0.22. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.22 - 0.25}{0.0399}$

Z = -0.75

Z = -0.75 has a pvalue of 0.2266

1 - 0.2266 = 0.7734

0.7734% probability that a random sample of size 118 exceeds 22%.

Sample of size 177:

$n = 177, so [tex]s = \sqrt{\frac{0.25*0.75}{118}} = 0.0325$

Probability of sample proportion above 22%.

This is 1 subtracted by the pvalue of Z when X = 0.22. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.22 - 0.25}{0.0325}$

Z = -0.92

Z = -0.92 has a pvalue of 0.1788

1 - 0.1788 = 0.8212

0.8212 = 82.12% probability that a random sample of size 177 exceeds 22%.

82.12% > 77.34%, so the sample proportion of size 177 is more likely to exceed 22%.

Answer from: Quest

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Answer from: Quest

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step-by-step explanation:

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