Question

Suppose a large shipment of stereos contained 18% defectives. If a sample of size 306 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 6%

Answer 1

99.36% probability that the sample proportion will differ from the population proportion by less than 6%

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 = (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 sample proportion p in a sample of size n, we have that the sampling distribution of the sample proportions has
`\mu = p, s = \sqrt{(p(1-p))/(n)}`.

In this question:

`n = 306, p = 0.18, \mu = 0.18, s = \sqrt{(0.18*0.82)/(306)} = 0.0220`.

What is the probability that the sample proportion will differ from the population proportion by less than 6%

This is the pvalue of Z when X = 0.18 + 0.06 = 0.24 subtracted by the pvalue of Z when X = 0.18 - 0.06 = 0.12. So

X = 0.24

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

By the Central Limit Theorem

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

`Z = (0.24 - 0.18)/(0.022)`

`Z = 2.73`

`Z = 2.73` has a pvalue of 0.9968

X = 0.12

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

`Z = (0.12 - 0.18)/(0.022)`

`Z = -2.73`

`Z = -2.73` has a pvalue of 0.0032

0.9968 - 0.0032 = 0.9936

99.36% probability that the sample proportion will differ from the population proportion by less than 6%