Question

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 a 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 size of 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision
Required:
a. For a test with a level of significance of 0.10, compute the critical value; set up the appropriate hypotheses and the calculate the p-value of the test
b. Will the null hypothesis will be rejected with a level of significance of 0.10? Explain why or why not?

Answer 1

Using the z-distribution, we have that:

a)

The hypothesis test is:

• $H_0: \mu = 7.4$

• $H_1: \mu < 7.4$

The p-value is of 0.0668.

The critical value is $z^{\ast} = -1.28$

b) Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

Item a:

At the null hypothesis, it is tested if the mean is of 7.4 minutes, that is:

$H_0: \mu = 7.4$

At the alternative hypothesis, it is tested if the mean is lower than 7.4 minutes, that is:

$H_1: \mu < 7.4$

We have the standard deviation for the population, hence, the z-distribution is used.

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• $\sigma$ is the standard deviation of the sample.

• n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36$.

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}$

$z = -1.5$

Using a z-distribution calculator, the p-value is of 0.0668.

Also using a calculator, considering a left-tailed test, as we are testing if the mean is less than a value, the critical value with a significance level of 0.1 is of $z^{\ast} = -1.28$.

Item b:

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918