D comment on the comparison between this result and

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Unformatted text preview: 0 b) Determine α = 0.05 (95% confidence interval) Critical value of T (d.f=4) =  ­2.78; 2.78 From table: b ( from table) =  ­0.3642 n SSE = ∑ ( H i − H i ) 2 = 8315 i =1 s= SSE 8315 = = 45.6 n−2 6−2 T = − 0.3642 − 0.0 =  ­0.197 45.6 / 614.83 So, the evidence fails to reject the null hypothesis. T is not in the rejection region. − 0.3642 − 0.0 c) T = =  ­0.197 45.6 / 614.83 P (T = ­0.197; d.f =4) = 0.85343> 0.05 use Excel: tdist(T ­value, d.f, 2 tails) So, the evidence fails to reject the null hypothesis 4/5 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 12 Hypothesis Testing 4) You are given the data set in Tabl...
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