A manufacturer of mosquito repellent claims that the average amount of time that its repellent is effective is at least 3.5 hr.A consumer group tests the claim using five randomly selected people.If the average time of effectiveness from the sample was 3.8 hr with a standard deviation of 0.76 hr, what is the critical value (CV), test statistic (TS), and decision for this test? (Use α = 0.025.)
A) CV = -2.7764; TS = 0.88; fail to reject H₀
B) CV = 2.7764; TS = 0.88; fail to reject H₀
C) CV = 2.5706; TS = 0.39; fail to reject H₀
D) CV = -2.5706; TS = -0.88; fail to reject H₀

Question

Quizplus · Accepted Answer

The critical value (CV) corresponds to the t-distribution with n-1 degrees of freedom (where n is the sample size) at a significance level of α = 0.025 for a one-tailed test. With 5 samples, the degrees of freedom are 4. Looking at a t-table or using statistical software for a one-tailed test with α = 0.025 and 4 degrees of freedom, the CV is approximately -2.7764. The test statistic (TS) is calculated using the formula TS = (x̄ - μ) / (s / √n), where x̄ is the sample mean, μ is the population mean under the null hypothesis, s is the sample standard deviation, and n is the sample size. Plugging in the given values, TS = (3.8 - 3.5) / (0.76 / √5) ≈ 0.88. Since the test statistic does not exceed the critical value in absolute terms, we fail to reject the null hypothesis, indicating that the data does not provide sufficient evidence to conclude that the average effectiveness of the repellent is less than 3.5 hours.

A Manufacturer of Mosquito Repellent Claims That the Average Amount