It is claimed that in a bushel of peaches,fewer than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. If α = 0.025,what will be the decision?
A)Accept the null.
B)Reject the null and conclude the defects are not greater than 10%.
C)Reject the null and conclude the defects are greater than 10%.
D)Fail to reject the null.

Question

Quizplus · Accepted Answer

The sample proportion of defective peaches is 50/400 = 0.125. The null hypothesis is that the proportion of defective peaches is less than 0.10. The alternative hypothesis is that the proportion of defective peaches is greater than 0.10. The test statistic is z = (0.125 - 0.10) / sqrt(0.10 * 0.90 / 400) = 2.5. The p-value is P(Z > 2.5) = 0.0062. Since the p-value is less than 0.025, we reject the null hypothesis and conclude that the proportion of defective peaches is greater than 0.10.

It Is Claimed That in a Bushel of Peaches,fewer Than