TABLE 9-4
A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an mean of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normally distributed with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision.
-Referring to Table 9-4, if the level of significance had been chosen as 0.05, the null hypothesis would be rejected.

Question

Quizplus · Accepted Answer

We need to calculate the test statistic using the formula:

z = (x̄ - μ) / (σ / √n)

Where x̄ is the sample mean, μ is the hypothesized population mean (7.4), σ is the population standard deviation (1.2), and n is the sample size (36).

z = (7.1 - 7.4) / (1.2 / √36) = -1.5

Looking up the critical value for a two-tailed test with α = 0.05 in Table 9-4, we find it to be ±1.96. Since -1.5 is not in the rejection region, we fail to reject the null hypothesis.

Therefore, we cannot conclude that the new anesthetic is more effective than the current one at the 0.05 level of significance.

TABLE 9-4 A Drug Company Is Considering Marketing a New Local Anesthetic