Answer:
0.0098 = 0.98% probability that the sample proportion will differ from the population proportion by greater than 0.04
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
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
Suppose the true proportion is 0.06.
This means that ![p = 0.06](https://tex.z-dn.net/?f=p%20%3D%200.06)
235 are sampled
This means that ![n = 235](https://tex.z-dn.net/?f=n%20%3D%20235)
Mean and standard deviation:
![\mu = p = 0.06](https://tex.z-dn.net/?f=%5Cmu%20%3D%20p%20%3D%200.06)
![s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.06*0.94}{235}} = 0.0155](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.06%2A0.94%7D%7B235%7D%7D%20%3D%200.0155)
What is the probability that the sample proportion will differ from the population proportion by greater than 0.04?
Proportion below 0.06 - 0.04 = 0.02 or above 0.06 + 0.04 = 0.1. Since the normal distribution is symmetric, these probabilities are equal, which means that we can find one of them and multiply by 2.
Probability the proportion is below 0.02.
p-value of Z when X = 0.02. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.02 - 0.06}{0.0155}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.02%20-%200.06%7D%7B0.0155%7D)
![Z = -2.58](https://tex.z-dn.net/?f=Z%20%3D%20-2.58)
has a p-value of 0.0049.
2*0.0049 = 0.0098
0.0098 = 0.98% probability that the sample proportion will differ from the population proportion by greater than 0.04