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 0.050.05. If 212212 are sampled, what is the probability that the sample proportion will differ from the population proportion by less than 0.030.03

Answer 1

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.

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 = (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 = (\sigma)/(√(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{(p(1-p))/(n)}`

In this question:

`p = 0.05, n = 212, \mu = 0.05, s = \sqrt{(0.05*0.95)/(212)} = 0.015`

What is the probability that the sample proportion will differ from the population proportion by less than 0.03?

This is the pvalue of Z when X = 0.03 + 0.05 = 0.08 subtracted by the pvalue of Z when X = 0.05 - 0.03 = 0.02. So

X = 0.08

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (0.08 - 0.05)/(0.015)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

X = 0.02

`Z = (X - \mu)/(s)`

`Z = (0.02 - 0.05)/(0.015)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.