An experiment on the side effects of pain relievers assigned arthritis patients

Question

An experiment on the side effects of pain relievers assigned arthritis patients to take one of several over-the-counter pain medications. Of the 427 patients who took one brand of pain reliever, 23 suffered some "adverse symptom." Does the experiment provide strong evidence that fewer than 9% of patients who take this medication have adverse symptoms? (a) H_0: p = and H_a p (b) The test statistic is (Use 2 decimal places (c) The p-value is (Use 4 decimal places (d) Therefore, we can conclude that (choose all that apply) The data does provide statistical evidence at the 0.05 significance level that fewer than 9% of arthritis patients taking the pain medication experience adverse symptoms. The data does provide statistical evidence at the 0.05 significance level that fewer than 9% of these 427 arthritis patients taking the pain medication experience adverse symptoms The data does not provide statistical evidence at the 0.05 significance level that fewer than 9% of arthritis patients taking the pain medication experience adverse symptoms. The data does provide statistical evidence at the 0.05 significance level that 5.39% of arthritis patients taking the pain medication experience adverse symptoms.

Explanation / Answer

Solution:-

x = 23, n = 427

p = 23/427 = 0.0539

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.09

Alternative hypothesis: P < 0.09

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.01385

z = (p - P) /

z = - 2.61

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 2.61. We use the Normal Distribution Calculator to find P(z < - 2.61) = 0.0045

Interpret results. Since the P-value (0.0045) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that fewer than 9% of the arthritis patients taking the pain medication experience adverse symptoms.

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