Exam1solac.Fall00

# Exam1solac.Fall00 - D E 14 Reject Ho because p-value is...

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Fall 2000 Exam 1 Solutions MSCI3710 No. Solution Version A Version C 1 The correct set of hypothesis for this test is Ho: 1 >= 393 Ha: 1 < 393 B A 2 Z = (X- μ )/( σ / n), (390-393)/(22 / 42) = -0.88 Variance is 484, so the standard deviation is 22 A D 3 p-value is 0.0202 and the alpha is 0.01. Therefore, Fail to reject the null hypothesis, conclude there is insufficient evidence of Ha. B D 4 t = (X- μ )/(s/ n), (438-443)/( 576 / 13) = -0.75 D D 5 Decision Rule: Reject Ho if | t | > 1.782 E B 6 Ho: μ 1= μ 2 Ha: μ 1 ≠μ 2 D C 7 Critical value : -1.96 and 1.96 B E 8 (2.61-2.38)/ (0.12^/50)+(0.14^/40)=8.25 C D 9 Reject Ho then what is the conclusion? B D 10 Ho: p 0.2 Ha: p < 0.2 D E 11 (0.22-0.2)/ {(0.2*0.8)/350}=1.069 A B 12 Look at the table on the front page of the text book. When Z= -2.45, Value is 0.4929 So, (0.5-0.4929)=0.0071 D A 13 From the table, p-value of the two tail is 0.0071

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Unformatted text preview: D E 14 Reject Ho because p-value is less than alpha. Then conclusion is that there is evidence of Ha. A C 15 Ho: p m 1 p w Ha: p m > p w A C 16 (89+116)/300=0.6833 A D 17 Men : 89/148=0.6Women:116/152=0.76 (0.6-0.76)/ √ {(0.68(0.32)/148+(0.68*0.32)/152}=-3.02 C D 18 From the table, when Z=-3.02, the value is 0.4987 So, (0.5-0.4987)=0.0013. then 1-0.0013=0.9987 E C 19 Ho: the mean time to deliver orders is the same for all three suppliers. B A 20 MS Between Groups/ MS within Groups 20.556/15.4476=1.3307 E A 21 From the table(A-7), alpha 0.01 df with 2,42 = 5.18 D B 22 P-value is greater than Alpha; therefore Reject Ho C A 23 42.2-31.2=11 and unequal B E 24 MS fund/ MS error 8.0593/0.9784=8.2370 A C 25 MS Columns/ MS error B E 8.3626/0.9784=8.5470...
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