Question

Consider a drug that is used to help prevent blood clots in certain patients. In clinical​ trials, among 6108 patients treated with this​ drug, 164 developed the adverse reaction of nausea. Use a 0.01 significance level to test the claim that 3​% of users develop nausea. Does nausea appear to be a problematic adverse​ reaction?

Identify the test statistic for this hypothesis test.

Answer 1

Lecture: For this study, one proportion z-test is used to test the claim. The problem deals with the concept of finding the p-value using the normal distribution as an approximation to the binomial distribution. Step by step solution is provided with formulas and explanation. The mean of the sampling distribution of sample proportions is, Mo =P The standard deviation of the sampling distribution of sample proportions is, p(1-P) п Here, n is the sample size, and p is the probability of success. The formula for the 2-test statistic is, р- да бр p-p p(1-P) n Here, p is the sample proportion. Note that the sampling distribution of sample proportions follows a normal distribution.

Solution: The null and alternative hypotheses are as follows: H:p=0.30 (the proportion of patients develop nausea is 3%) H : p=0.00 (Claim) (the proportion of patients develop nausea is not 3%) Here, p is the population proportion. Let the number of patients developed the adverse reaction of nausea be x = 164. Let the total number of patients in a sample ben = 6108, The sample proportion of patients developed the adverse reaction of nausea is, x n 164 = 6108 = 0.026850 Under the null hypothesis, the test statistic value is, P p P(1-P) n 0.026850-0.30 10.30(1-0.30) 6986 = -1.4431419 = -1.44

Find the p-value. The alternative hypothesis contains not equal to symbol. Now, it is necessary to find the two-tailed p-value corresponding to the test statistic value of -=-1.44. Using the excel function formula, the left-tailed p-value is as follows: p-value = 2P(:<-3.77) = 2(=NORMSDIST(-1.44)] = 2(0.0749337) = 0.149867399 = 0.1499 Therefore, the p-value is 0.1499. Let the level of significance be a = 0.01. Conclusion: The p-value is greater than the significance level of 0.01. It fails to reject the null hypothesis. Therefore, it can be concluded that there is no sufficient evidence to support the claim that 3% of users develop nausea. So, nausea do not appear to be a problematic adverse reaction.

Note: Technology help For quick reference, the Minitab output for the one proportion z-test is as follows: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Sample p 99% CI for p 6108 164 0.026850 (0.021522, 0.032178) Test Null hypothesis Ho: p = 0.03 Alternative hypothesis H:P 0.03 Z-Value P-value -1.44 0.14898 Note: Answer verified.