Question

Juries should have the same racial distribution as the surrounding communities. According to the U.S. Census Bureau, 18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American. How likely is the resulting sample proportion to be between 0.066 and 0.294 (i.e., 6.6% to 29.4% African American)

Answer 1

Answer:

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 $s = \sqrt{\frac{p(1-p)}{n}}$

18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.

This means that $p = 0.18, n = 100$

So, by the Central Limit Theorem:

$\mu = 0.18, s = \sqrt{\frac{0.18*0.82}{100}} = 0.0384$

How likely is the resulting sample proportion to be between 0.066 and 0.294?

This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So

X = 0.294

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

By the Central Limit Theorem

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

$Z = \frac{0.294 - 0.18}{0.0384}$

$Z = 2.97$

$Z = 2.97$ has a pvalue of 0.9985

X = 0.066

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

$Z = \frac{0.066 - 0.18}{0.0384}$

$Z = -2.97$

$Z = -2.97$ has a pvalue of 0.0015

0.9985 - 0.0015 = 0.997

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294