Exam1solbd.Fall00

Exam1solbd.Fall00 - conclusion is that there is...

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Fall 2000 Exam 1 Solutions MSCI3710 No. Solution Version B Version D 1 The correct set of hypothesis for this test is Ho: 1 >= 483 Ha: 1 < 483 B E 2 Z = (X- μ )/( σ / n), (490-483)/(22 / 40) = 2.01 Variance is 484, so the standard deviation is 22 A B 3 p-value is 0.04 and the alpha is 0.05. Therefore, Reject the null hypothesis, conclude there is sufficient evidence of Ha. C E 4 t = (X- μ )/(s/ n), (538-545)/( 784 / 21) = -1.15 A A 5 Decision Rule: Reject Ho if | t | > 1.325 E A 6 Ho: μ 1= μ 2 Ha: μ 1 ≠μ 2 D A 7 Critical value : -2.58 and 2.58 A E 8 (2.78-2.56)/ (0.15^/60)+(0.18^/50)=6.88 C E 9 Reject Ho There is sufficient evidence of Ha B A 10 Ho: p 0.19 Ha: p < 0.19 D B 11 (0.22-0.19)/ {(0.19*0.81)/440}=1.75 A E 12 Look at the table on the front page of the text book. When Z= -2.885, Value is 0.4980 So, (0.5-0.4980)=0.0020 D A 13 From the table, p-value of the two tail is 0.0142 D C 14 Reject Ho because p-value is greater than alpha. Then
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Unformatted text preview: conclusion is that there is insufficient evidence of Ha. B D 15 Ho: p m 1 p w Ha: p m > p w A C 16 (124+131)/400=0.6375 A C 17 Men : 124/218=0.57 Women:131/182=0.72 (0.57-0.72)/ √ {(0.64(0.36)/218+(0.64*0.36)/182}= -3.13 C E 18 From the table, when Z=-3.13, the value is 0.4991 So, (0.5-0.4991)=0.0009. then 1-0.0009=0.9991 E A 19 Ho: the mean time to deliver orders is the same for all three suppliers. B A 20 MS Between Groups/ MS within Groups 16.4667/14.1397=1.1646 E A 21 From the table(A-7), alpha 0.01 df with 2,42 = 5.18 D C 22 P-value is greater than Alpha; therefore Fail to Reject Ho C A 23 42.2-31.2=11 and unequal B E 24 MS fund/ MS error 13.6604/2.7487=4.9698 A B 25 MS Columns/ MS error 19.0937/2.7487=6.9464 B C...
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This note was uploaded on 03/09/2011 for the course DSC 3710 taught by Professor Staff during the Spring '09 term at North Texas.

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Exam1solbd.Fall00 - conclusion is that there is...

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