Question

Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

Answer 1

0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

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 = (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 = (\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)}`

Suppose a large shipment of televisions contained 9% defectives

This means that
`p = 0.09`

Sample of size 393

This means that
`n = 393`

Mean and standard deviation:

`\mu = p = 0.09`

`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.09*0.91)/(393)} = 0.0144`

What is the probability that the sample proportion will differ from the population proportion by less than 3%?

Proportion between 0.09 - 0.03 = 0.06 and 0.09 + 0.03 = 0.12, which is the p-value of Z when X = 0.12 subtracted by the p-value of Z when X = 0.06.

X = 0.12

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

By the Central Limit Theorem

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

`Z = (0.12 - 0.09)/(0.0144)`

`Z = 2.08`

`Z = 2.08` has a p-value of 0.9812

X = 0.06

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

`Z = (0.06 - 0.09)/(0.0144)`

`Z = -2.08`

`Z = -2.08` has a p-value of 0.0188

0.9812 - 0.0188 = 0.9624

0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%