Exam1Sp02DKey

Exam1Sp02DKey - the book. p/2<0.05, so...

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Spring 2002 Exam 1 Solutions MSCI3710 No. Solution Version D 1 The correct set of hypothesis for this test is Ho: μ = 7400 Ha: μ 7400 E 2 When population standard deviation is known and the sample size is large, use Z test D 3 Z= (X- μ )/( σ / n)=(7720-7400)/(1050 / 30) = 1.67 D 4 Z*=1.67 > Z (alpha/2)=1.645, so Reject Ho, which means the average is not equal to $7400 D 5 Z (alpha/2)=1.646. Because this is the two-tail test. The rejection area should be to the left of Z=-1.645, and to the right of Z=1.645 B 6 If Z* = 1.645, p=0.05*2=0.1 A 7 The correct set of hypothesis for this test is Ho: μ < 195 Ha: μ >1195 A 8 Z= (X- μ )/(s/ n)=(210.66-195)/(35.2 / 20) = 1.99 B 9 1.729 < T* (alpha, 19) = 1.85 <2.093. We can get .025 <p< .05 C 10 Because P=.03 > alpha=.01, Fail to reject Ho. So μ < 195 A 11 S p 2 = (n1-1) S 1 2 + (n2-1) S 2 2 = 14*(4.5 2 ) + 12*(3.4 2 ) = 16.2 n 1 + n 2 – 2 26 B 12 T (alpha/2, df)= t (.025,26) =2.056 C 13 t = X 1 - X 2 = 27.5-23.6 = 2.31 S p * 1/n 1 + 1/n 2 4.45 * 1/15+ 1/13 D 14 T (p/2, df)=5.67, df=n1+n2-2=26. Look at the table in the front of
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Unformatted text preview: the book. p/2&lt;0.05, so p&lt;0.01 E 15 The correct set of hypothesis for this test is Ho: p 0.70 Ha: p&lt;0.70 C 16 N= Z 2 *p(het)(1-p(het)) = 2.575 2 *(124/190)(1-124/190) = 237 E 2 0.08 2 E 17 Z = p(het)-po = 124/190-.7 = -1.43 po(1-po)/n .7*.3/190 A 18 P(het)+ Z p(het)*(1-p(het))/(n-1). 124/190+ 2.575 124/190)*(1-124/190)/(190-1). So it is from 0.563 to 0.742 D 19 This is one-tail test. So the reject area should be Z* (.01)&lt; -2.33 B 20 F = MS Columns = 0.641 = 10.02 MS Error 0.064 C 21 The correct set of hypothesis for this test is Ha: At lease one pair of 1, 2, 3 is different E 22 F*=1.7 &lt; F(0.05, 2, 8)=4.46, so Fail to reject Ho D 23 SS Row = 2.251 * 4 = 9.00 B 24 F* = MS Rows = 2.251 = 35.17 MS Error 0.064 A 25 F alpha, 4, 8 = 16.50 &gt; F .01,4,8 = 7.01 p &lt; .01 E...
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Exam1Sp02DKey - the book. p/2&amp;amp;lt;0.05, so...

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