Question

A buyer for a grocery chain inspects large truckloads of apples to determine the proportion p of apples in the shipment that are rotten. She will only accept the shipment if there is clear evidence that this proportion is less than 0. 06 she selects a simple random sample of 200 apples from the over 20000 apples on the truck to test the hypotheses h0: p = 0. 06, ha: p < 0. 6. The sample contains 9 rotten apples. The p-value of her test is

Answer 1

approximately 0.0002 (or 0.02%).

Explanation:

To find the p-value, we need to calculate the probability of getting a sample proportion of 9/200 or less assuming the null hypothesis is true (i.e. assuming that the true proportion of rotten apples in the population is 0.06).

We can use a normal approximation to the binomial distribution, since n = 200 is large enough and 200(0.06) = 12 is greater than 10. The test statistic is:

z = (x - np) / sqrt(np(1-p))

where x is the number of rotten apples in the sample (9), n is the sample size (200), and p is the hypothesized proportion (0.06).

Substituting these values, we get:

z = (9 - 200(0.06)) / sqrt(200(0.06)(0.94)) ≈ -4.07

The p-value is the probability of getting a z-value of -4.07 or less, which we can find using a standard normal distribution table or calculator. This probability is approximately 0.0002.

Since the p-value is very small (much less than 0.05), we reject the null hypothesis and conclude that there is clear evidence that the proportion of rotten apples in the shipment is less than 0.06. The buyer can accept the shipment.