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)? (Hint: Use the sampling distribution of the sample proportion in this case.)

Answer 1

Answer:

Step-by-step explanation:

For the given scenario, we are given p = 20% = 0.20

The standard deviation for the sampling distribution of proportion is given as below:

SD = sqrt(pq/n)

Where, p = 0.20, q = 1 – p = 1 – 0.20 = 0.80 and n = 400

SD = sqrt(0.20*0.80/400) = sqrt(0.0004) = 0.02

Now, by using empirical rule (68-95-99.7 rule)

1SD = 1*0.02 = 0.02

About 68% chance that the resulting sample proportion will be within 0.02 of the true proportion.

2SD = 2*0.02 = 0.04

About 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

3SD = 3*0.02 = 0.06

About 99.7% chance that the resulting sample proportion will be within 0.06 of the true proportion.

So, correct answer:

There is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

Answer 2

Answer:

95.44% probability the resulting sample proportion is 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 normally distributed samples are 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 the sampling distribution of the sample proportion in sample of size n, the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.2, n = 400$

So

$\mu = 0.2, s = \sqrt{\frac{0.2*0.8}{400}} = 0.02$

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.

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

95.44% probability the resulting sample proportion is within .04 of the true proportion.