Critical value: z = 2.33. Fail to reject null hypothesis. There is not sufficient evidence to support the claim
that more than half of all those using the drug experience relief.
-A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawn equipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha. Find the
P-value for a test to support the claim that the proportion with lawn mowers in Omaha is higher than 65%.
Among 497 randomly selected homes in Omaha, 340 had one or more lawn mowers.
A) 0.1118
B) 0.0559
C) 0.0505
D) 0.0252

Question

Quizplus · Accepted Answer

We can use the one-sample proportion z-test.

The null hypothesis is that the proportion with lawn mowers in Omaha is 65% or less. The alternative hypothesis is that it is greater than 65%.

We can calculate the test statistic as:

z = (340/497 - 0.65) / sqrt(0.65 * 0.35 / 497) = 1.98

Using a standard normal distribution table, the p-value is approximately 0.025 (or 0.05 for a two-tailed test).

Since the p-value is less than 0.05 (or 0.025 for a one-tailed test), we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the proportion with lawn mowers in Omaha is higher than 65%. The best choice is B.

Critical Value: Z = 2