Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Suppose both populations of hellbenders are known to have ranavirus infections at a rate of 25%.
This means that [tex]\mu = p = 0.25[/tex]
Sample of size 118:
[tex]n = 118, so [tex]s = \sqrt{\frac{0.25*0.75}{118}} = 0.0399[/tex]
Probability of sample proportion above 22%.
This is 1 subtracted by the pvalue of Z when X = 0.22. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.22 - 0.25}{0.0399}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] 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:
[tex]n = 177, so [tex]s = \sqrt{\frac{0.25*0.75}{118}} = 0.0325[/tex]
Probability of sample proportion above 22%.
This is 1 subtracted by the pvalue of Z when X = 0.22. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.22 - 0.25}{0.0325}[/tex]
[tex]Z = -0.92[/tex]
[tex]Z = -0.92[/tex] 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%.
The sample proportion that is more likely to exceed 22% is; the sample proportion with a sample size of 177
Central Limit Theorem
We are given;
Population proportion; p = 25% = 0.25
Now, in central limit theorem, the formula for standard deviation is;
σ = √(p(1 - p)/n)
For a sample size of 118, we have;
σ = √(0.25(1 - 0.25)/118)
σ = √0.001588983
σ = 0.03986
Similarly, for a sample size of 177, we have;
σ = √(0.25(1 - 0.25)/177)
σ = √0.001059322
σ = 0.032547
Since we want to find which of the two sample proportions will likely exceed 22% which is p₀ = 0.2, we will use the formula;
z = (p₀ - p)/σ
For sample size of 118, we have;
z = (0.22 - 0.25)/0.03986
z = -0.75
From online p-value from z-score calculator, we have;
p-value = 0.7734 or 77.34%
For sample size of 177, we have;
z = (0.22 - 0.25)/0.032547
z = -0.92
From online p-value from z-score calculator, we have;
p-value = 0.8212 or 82.12%
The p-value of the sample of 177 is higher than the p-value for the sample of 118 and so we can say that the sample proportion with a sample size of 177 is more likely to exceed 22%
Read more about Central Limit theorem at; https://brainly.com/question/25800303
