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.
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