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Quiz 5 Key MA 223 √ 1. The random variable Z = (x̄ − µ)/(σ/ n) is approximately standard normal, since n = 50 ≥ 30. Also, we may replace σ = s under these conditions. With x̄ = 2.21 this leads to 2.21 − µ √ < 1.96 −1.96 < 0.03/ 50 with 95 percent confidence. A bit of rearrangement yields 2.2017 < µ < 2.218 as the 95 percent confidence interval. 2. Repeat the last part but leave n undefined (and note we “know” that σ ≈ 0.03). The confidence interval would be √ √ 2.21 − (1.96)(0.03)/ n < µ < 2.21 + (1.96)(0.03)/ n. √ √ √ The length of this interval is (2)(1.96)(0.03)/ n, or 0.1176/√ n. We thus want 0.1176/ n ≤ 0.01. Divide both sides by 0.1176 and reciprocate to get n ≥ 11.76. Square to find that we need n ≥ 139 if we round up. 1