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CEE 202 Homework 12 Solution

# 996 one tail so p p1p2 0 0004 04 probability that

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Unformatted text preview: 2 = Nuclear power a) H0: p1 ­p2 ≤ 0 b) Critical value approach: n1 = 1012 n2 = 870 ˆ p1 = 0.63, ˆ q1 = 0.37 ˆ q2 = 0.43 ˆ p2 = 0.57 ˆ ˆ p1 − p2 = 0.63 − 0.57 = 0.060 ˆˆ ˆˆ p1q1 p2 q2 + = n1 n2 0.63 ⋅ 0.37 0.57 ⋅ 0.43 + = 0.0226 1012 870 ˆ ˆ ( p1 − p2 ) − ( p1 − p2 ) 0.060 = = 2.654 0.0224 ˆˆ ˆˆ p1q1 p2 q2 + n1 n2 • Determine α = 0.05 (95% confidence interval) Critical value of Zcrit = 1.64 So, Z > Zcrit , p1 ­p2 > 0 • So the null hypothesis can be rejected. The value of Z lies in the rejection region. We have greater than 95% confidence that p1>p2. ˆ ˆ ( p − p2 ) − ( p1 − p2 ) 0.060 c) z = 1 = = 2.651 0.0224 ˆˆ ˆˆ p1q1 p2 q2 + n1 n2 For z =2.651, area under the normal curve = 0.996 (one tail) So, P (p1 ­p2 =0) = 0.004, 0.4% probability that the null hypothesis is true— very small. • So we reject the null hypothesis with high confidence. z= 3/5 CEE 20...
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