3. A study of "adverse symptoms" in users of over-the-counter paine relievers: acetaminophen and ibuprofen. In all, 650 subjects took random to one of two common pain rese acetaminophen, and 44 experienced some adverse symptom. Of the 347 subjects who took ibuprofen, 49 had an adverse symptom. (a) Does the data provide convincing evidence that the two pain relievers differ in the proportion of people who experience an adverse symptom? Support your conclusion with a test of significance. Use \( \alpha=0.05 \). (b) Find the margin of error for a \( 95 \% \) confidence interval for the difference in the proportions of people who experience adverse reactions to these two mediations. You need not carry out all the steps in constructing a confidence interval. (c) In parts (a) and (b) above, you should have used two different formulas for estimating the standard deviation of the randomization distribution of \( \hat{p}_{1}-\hat{p}_{2} \). Explain why it makes sense to do this.

We are given two independent groups: • Acetaminophen: n₁ = 650 with x₁ = 44 adverse events, so p̂₁ = 44⁄650 ≈ 0.0677. • Ibuprofen: n₂ = 347 with x₂ = 49 adverse events, so p̂₂ = 49⁄347 ≈ 0.1412. We wish to test and estimate the difference in the proportions experiencing an adverse symptom. ────────────────────────────── Part (a). Test of Significance Step 1. State the hypotheses:   H₀: p₁ = p₂  (no difference)   Hₐ: p₁ ≠ p₂  (two‐sided alternative) Step 2. Compute the pooled proportion under H₀:   p̂ = (x₁ + x₂)⁄(n₁ + n₂) = (44 + 49)⁄(650 + 347) = 93⁄997 ≈ 0.0933. Step 3. Compute the standard error (SE) for the difference under H₀:   SE = √[p̂ (1 − p̂) (1/n₁ + 1/n₂)] Calculate:   1/n₁ = 1/650 ≈ 0.001538, 1/n₂ = 1/347 ≈ 0.002882.   Sum = 0.001538 + 0.002882 = 0.004420.   Now, p̂ (1 − p̂) = 0.0933 × 0.9067 ≈ 0.0846.   Thus, SE ≈ √(0.0846 × 0.004420) ≈ √(0.000374) ≈ 0.01934. Step 4. Compute the test statistic:   z = (p̂₁ − p̂₂)⁄SE = (0.0677 − 0.1412)⁄0.01934 ≈ (−0.0735)⁄0.01934 ≈ −3.80. Step 5. Determine the p-value: Since the test is two‐sided, the p-value is 2P(Z < −3.80). Looking up the standard normal distribution (or using software) shows that this p-value is very small (p < 0.001). Step 6. Conclusion: Because p < 0.05, we reject H₀. The data provide convincing evidence that the two pain relievers differ in the proportion of people experiencing an adverse symptom. ────────────────────────────── Part (b). 95% Confidence Interval for (p₁ − p₂) For the confidence interval we do not assume p₁ = p₂, so we use the unpooled standard error. Step 1. Compute the unpooled standard error:   SE_CI = √[(p̂₁ (1 − p̂₁))/n₁ + (p̂₂ (1 − p̂₂))/n₂]. Calculations:  For acetaminophen:   p̂₁ (1 − p̂₁) = 0.0677 × 0.9323 ≈ 0.0631,   Divide by n₁: 0.0631⁄650 ≈ 0.0000971.  For ibuprofen:   p̂₂ (1 − p̂₂) = 0.1412 × 0.8588 ≈ 0.1213,   Divide by n₂: 0.1213⁄347 ≈ 0.0003497.  Summing: 0.0000971 + 0.0003497 ≈ 0.0004468.  Thus, SE_CI ≈ √(0.0004468) ≈ 0.02114. Step 2. For 95% confidence, the critical value is approximately z* = 1.96. The margin of error (ME) is:   ME = z* × SE_CI ≈ 1.96 × 0.02114 ≈ 0.0414. Thus, the 95% confidence interval for (p₁ − p₂) would be:   (p̂₁ − p̂₂) ± 0.0414   = (0.0677 − 0.1412) ± 0.0414   ≈ (−0.0735 ± 0.0414). This results in an interval of approximately (−0.1149, −0.0321). (The margin of error is 0.0414.) ────────────────────────────── Part (c). Explanation of the Two Different Standard Error Formulas In testing the null hypothesis (p₁ = p₂), we assume that the two population proportions are equal. Under this assumption, pooling the data to produce a combined proportion (p̂) gives a better estimate of the common proportion. This pooled proportion is then used in calculating the standard error for the test statistic, since the null hypothesis provides a “model” in which a single parameter describes both groups. For the confidence interval on (p₁ − p₂), however, we are estimating the difference between two proportions without assuming they are equal. Therefore, we must use estimates from each sample separately to capture the variability in the two independent estimates. That is why we use the unpooled formula for the standard error when constructing the confidence interval. ────────────────────────────── Summary Answers (a) The two‐proportion z–test statistic is approximately –3.80 with p < 0.001, so there is convincing evidence at the α = 0.05 level that the proportions of adverse reactions differ between acetaminophen and ibuprofen. (b) The margin of error for a 95% confidence interval is approximately 0.0414 (using the unpooled standard error). (c) For the significance test we use a pooled standard error because we assume under H₀ that the proportions are equal. In contrast, for the confidence interval, we do not assume equal proportions and therefore use the unpooled standard error that incorporates the variability from each individual sample. These results comprehensively address the problem's parts.