Question

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion to be within .04 of the true proportion (i.e., between .16 and .24)

Answer 1

Answer:

0.9544 = 95.44% probability of the resulting sample proportion being within .04 of the true proportion

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 z-score 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 p-value, 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 $s = \sqrt{\frac{p(1-p)}{n}}$

20% of the residents in a certain state support an increase in the property tax. Sample of 400.

This means that $p = 0.2, n = 400$

Mean and standard deviation:

Mean $\mu = p = 0.2$

Standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2*0.8}{400}} = 0.02$

How likely is the resulting sample proportion to be within .04 of the true proportion (i.e., between .16 and .24)?

This is the pvalue of Z when X = 0.24 subtracted by the pvalue of Z when X = 0.16. So

X = 0.24

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

By the Central Limit Theorem

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

$Z = \frac{0.24 - 0.2}{0.02}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 0.16

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

$Z = \frac{0.16 - 0.2}{0.02}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability of the resulting sample proportion being within .04 of the true proportion