What is the probability that 20% (.2) or less of the population of toys is defective?

Question

A quality control worker in a Beanie Babies factory sampled 150 toys and found that 25 were defective.

What is the probability that 20% (.2) or less of the population of toys is defective?

Answer 1

We are given,

sample size, n=150, number of defectives in the sample, x=25

the sample proportion p-hat = x/n = 25/150 =0.1667

We have to find the probability that the true population proportion, p .≤ 2, when the sample proportion is p-hat is 1.667. Under normal approximation for proportions, p-hat follows a normal distribution with mean=p, and SE= sqrt(p*(1-p)/n) which is the standard error for proportions.

Let's compute the SE = (sqrt(0.2*(1-0.2)/150)=0.0327

The test statistics, z=(p-hat-p)/SE = (0.1667-0.2)/0.327 = -1.02

Probability p(z ≤-1.02)=0.154 ( from the z-table or use graphing calc or other technology)

Or, P(p ≤ 0.2)=0.154=about 15.4%

Hence, there is 15.4% probability that 20% or less of the population of toys is defective.