Question

A statistician calculates that 7% of Americans are vegetarians. If the statistician is correct, what is the probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%

Answer 1

Answer:

0.0182 = 1.82% probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%.

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 establishes 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 $s = \sqrt{\frac{p(1-p)}{n}}$

A statistician calculates that 7% of Americans are vegetarians.

This means that $p = 0.07$

Sample of 403 Americans

This means that $n = 403$

Mean and standard deviation:

$\mu = p = 0.07$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.07*0.93}{403}} = 0.0127$

What is the probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%?

Proportion below 7 - 3 = 4% or above 7 + 3 = 10%. Since the normal distribution is symmetric, these probabilities are equal, which means that we find one of them, and multiply by 2.

Probability the proportion is below 4%

p-value of Z when X = 0.04.

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

By the Central Limit Theorem

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

$Z = \frac{0.04 - 0.07}{0.0127}$

$Z = -2.36$

$Z = -2.36$ has a p-value of 0.0091

2*0.0091 = 0.0182

0.0182 = 1.82% probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%.