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Exam1Sp02CKey

# Exam1Sp02CKey - the book 3.86>2.763 p<.005*2=.01 E 15...

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Spring 2002 Exam 1 Solutions MSCI3710 No. Solution Version C 1 The correct set of hypothesis for this test is Ho: μ = 7500 Ha: μ 7500 C 2 When population standard deviation is known and the sample size is large, use Z test A 3 Z= (X- μ )/( σ / n)=(7750-7500)/(975 / 36) = 1.54 D 4 Z ( alpha/2)=1.645. Because this is the two-tail test. The reject area should be to the left of Z=-1.645, and to the right of Z=1.645 B 5 Z*=2.60 > Z ( alpha/2)=1.645, so Reject Ho, which means the average is not equal to \$7500 C 6 If Z* = 1.645, p=0.05*2=0.10 A 7 The correct set of hypothesis for this test is Ho: μ < 200 Ha: μ >1200 D 8 Z= (X- μ )/( s / n)=(210.66-200)/(35.2 / 20) = 1.35 B 9 1.328 < T* (alpha, 19) = 1.40 <1.729. We can get .05 <p< .1 B 10 Because P=.07 < alpha=.1, Reject Ho. So μ > 200 A 11 S p 2 = (n1-1) S 1 2 + (n2-1) S 2 2 = 15*(4.4 2 ) + 13*(3.3 2 ) = 15.4 n 1 + n 2 – 2 28 B 12 T (alpha/2, df)= t (.025,28) = 2.048 C 13 t = X 1 - X 2 = 27.4-23.5 = 2.52 S p * 1/n 1 + 1/n 2 4.24 * 1/16+ 1/14 D 14 T (p/2, df)=3.86, df=n1+n2-2=28. Look at the table in the front of

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Unformatted text preview: the book. 3.86>2.763, p<.005*2=.01 E 15 The correct set of hypothesis for this test is Ho: p ≥ 0.60 Ha: p<0.60 C 16 Z = p(het)-po = 99/180-.6 = -1.37 √ po(1-po)/n √ .6*.4/180 A 17 P(het)+ Z √ p(het)*(1-p(het))/(n-1). 99/180+ 2.575 √ 99/180)*(1-99/180)/(180-1). So it’s from 0.454 to 0.6458 D 18 This is one-tail test. So the reject area should be Z* (.01)< -2.33 B 19 N= Z 2 *p(het)(1-p(het)) = 2.575 2 *(99/180)(1-99/180) = 337 E 2 0.07 2 E 20 F = MS Columns = 0.685 = 15.22 MS Error 0.045 A 21 The correct set of hypothesis for this test is Ha: At lease one pair of μ 1, μ 2, μ 3 is different E 22 F*=1.6 < F(0.05, 2, 8)=4.46, so Fail to reject Ho D 23 SS Row = 2.336 * 4 = 9.34 B 24 F* = MS Rows = 2.336 = 51.91 MS Error 0.045 A 25 F alpha, 4, 8 =14.5 > F .01,4,8 = 7.01 p <.01 E...
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Exam1Sp02CKey - the book 3.86>2.763 p<.005*2=.01 E 15...

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