Question

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.06. If 235 are sampled, what is the probability that the sample proportion will differ from the population proportion by greater than 0.04

Answer 1

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

Suppose the true proportion is 0.06.

This means that $p = 0.06$

235 are sampled

This means that $n = 235$

Mean and standard deviation:

$\mu = p = 0.06$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.06*0.94}{235}} = 0.0155$

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

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

By the Central Limit Theorem

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

$Z = \frac{0.02 - 0.06}{0.0155}$

$Z = -2.58$

$Z = -2.58$ 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