Question

Hypothesis Testing Worksheet 1. The cold medicine Dozenol list 600 milligrams of acetaminophen per fluid Ounce as an active ingredient. The Food and Drug Administration tests 65 one-ounce samples of the medicine and finds that the mean amount of acetaminophen for the sample is 589 milligrams with a standard deviation of 21 milligrams. Test the claim of the FDA that the medicine does not contain the required amount of acetaminophen at the level of significance 0.05. 2. The Medassist Pharmaceutical Company bottles aspirin tablets in batches of 500 tablets. It is imperative that each bottle have at least 500 tablets. As a regulator for the FDA, you want to test the claim that consumers are getting less than the advertised 500 tablets per bottle. You select a random sample of 200 bottles and find that they have a mean of 499 tablets and a standard deviation of tablets At a level of significanceof 0.05, what would you conclude about the hypothesis test?

Answer 1

Answer 1)

The provided sample mean is X = 589 and the sample standard deviation is s = 21, and the sample size is n = 65. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p = 600 Ha: u 7600 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two- tailed test is te = 1.998. The rejection region for this two-tailed test is R = {t: t > 1.998} (3) Test Statistics The t-statistic is computed as follows: X - HO 589 - 600 ts/m 21/v65 = -4.223 (4) Decision about the null hypothesis Since it is observed that t = 4.223 >te = 1.998, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0001, and since p = 0.0001 <0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is different than 600, at the 0.05 significance level.

Thus, there is enough evidence to support the FDA claim that the medicine does not contain the required amount of acetaminophen at 0.05 significance level.

Answer 2)

The provided sample mean is X = 499 and the sample standard deviation is s = 7, and the sample size is n = 200. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 2 500 Ha: u <500 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is te = -1.653. The rejection region for this left-tailed test is R=t:t < -1.653 (3) Test Statistics The t-statistic is computed as follows: _X-MO ks vn 499 - 500 7/ 200 -= -2.02 (4) Decision about the null hypothesis Since it is observed that t = -2.02 <te= -1.653, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0223, and since p=0.0223<0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean is less than 500, at the 0.05 significance level.

Thus, at 0.05 significance level, there is enough evidence to claim that consumers are getting less than the advertised 500 tablets per bottle.