Sample Proportion Question

Question

According to a 2009 Reader's Digest article, people throw away about 7% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 100 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.18?

Answer 1

We are given the following information:

p₀ = 0.07 (true proportion)

n = 100 grocery shoppers (sample size)

p-hat = 0.18 (sample proportion)

We want to check to see if the normal approximation can be used.

np ≥ 5 ⇒ 100*0.07 = 7 ≥ 5

n(1 - p) = nq ≥ 5 ⇒ 100*0.93 = 93 ≥ 5 (Note: q is the probability that a shopper didn't throw away food.)

Both conditions are met. Now, lets calculate the mean and standard deviation.

Mean: μ = p₀ = 0.07 (this information was given)

Standard Deviation: σ = √p₀(1 - p₀) / n = √(0.07)(0.93)/100 = 0.2551470164434614743903999267847

Find the probability using the z-score formula (this can be found in the textbook or online).

z = (p-hat - µ) / σ = (0.18 - 0.07) / 0.255 = 0.43

P(z < 0.43) = 0.6664

The probability that the sample proportion does not exceed 0.18 is 0.6664.