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Unformatted text preview: Section 98 993 1. The variable of interest is breakdowns among shift. 2. H0: Breakdowns are independent of shift. 3. H1: Breakdowns are not independent of shift. 4. = 0.05 5. The test statistic is: = 2 0 r c (O ij  Eij ) 2 i =1 j =1 Eij 6. The critical value is .05 , 6
2 7. = 12.592 2 The calculated test statistic is 0 = 11.65 8. 2 > 2.05,6 , do not reject H0 and conclude that the data provide insufficient evidence to claim that 0 / 0 machine breakdown and shift are dependent at = 0.05. Pvalue = 0.070 (using Minitab) 994 1. The variable of interest is calls by surgicalmedical patients. 2. H0:Calls by surgicalmedical patients are independent of Medicare status. 3. H1:Calls by surgicalmedical patients are not independent of Medicare status. 4. = 0.01 5. The test statistic is: = 2 0 r c (O ij  Eij ) 2 i =1 j =1 Eij = 6.637 2 7. The calculated test statistic is 0 = 0.033
2 6. The critical value is .01,1
2 2 0 > 0.01,1 / 8. , do not reject H0 and conclude that the evidence is not sufficient to claim that surgical medical patients and Medicare status are dependent. Pvalue = 0.85 946 995. 1. The variable of interest is statistics grades and OR grades. 2. H0: Statistics grades are independent of OR grades. 3. H1: Statistics and OR grades are not independent. 4. = 0.01 5. The test statistic is: = 2 0 r c (O ij  Eij ) 2 i =1 j =1 Eij 6. The critical value is
2 2 0 > 0.01,9 = 21.665 2 7. The calculated test statistic is 0 = 25.55
2 .01, 9 8. Therefore, reject H0 and conclude that the grades are not independent at = 0.01. Pvalue = 0.002 996 1. The variable of interest is characteristic among deflections and ranges. 2. H0: Deflection and range are independent. 3. H1: Deflection and range are not independent. 4. = 0.05 5. The test statistic is: = 2 0 r c (O ij  Eij ) 2 i =1 j =1 Eij 6. The critical value is
2 2 0 > 0.05, 4 / = 9.488 2 7. The calculated test statistic is 0 = 2.46
2 0.05 , 4 8. , do not reject H0 and conclude that the evidence is not sufficient to claim that the data are not independent at = 0.05. Pvalue = 0.652 997. 1. The variable of interest is failures of an electronic component. 2. H0: Type of failure is independent of mounting position. 3. H1: Type of failure is not independent of mounting position. 4. = 0.01 5. The test statistic is: = 2 0 r c (O ij  Eij ) 2 i =1 j =1 Eij = 11.344 2 7. The calculated test statistic is 0 = 10.71
2 6. The critical value is 0.01, 3
2 2 0 > 0.01,3 / 8. , do not reject H0 and conclude that the evidence is not sufficient to claim that the type of failure is not independent of the mounting position at = 0.01. Pvalue = 0.013 998 1. The variable of interest is opinion on core curriculum change. 2. H0: Opinion of the change is independent of the class standing. 3. H1: Opinion of the change is not independent of the class standing. 4. = 0.05 5. The test statistic is: 947 02 = i =1 j =1 r c (O ij  Eij ) 2 Eij 6. The critical value is 0.05 , 3
2 = 7.815 2 7. The calculated test statistic is .0 = 26.97 .
2 2 0 >>> 0.05,3 , reject H0 and conclude that the opinions on the change are not independent of class 8. standing. Pvalue 0 999 a) 1. The variable of interest is successes. 2. H0: successes are independent of size of stone. 3. H1: successes are not independent of size of stone. 4. = 0.05 5. The test statistic is: = 2 0 r c (O ij  Eij ) 2 i =1 j =1 Eij 6. The critical value is .05 ,1 = 3.84
2 7. The calculated test statistic 8. 2 0 = 13.766 with details below. >
2 0 2 0.05,1 , reject H0 and conclude that there is enough evidence to claim that number of successes and the stone size are not independent.
1 2 All 55 25 80 66.06 13.94 80.00 2 234 36 270 222.94 47.06 270.00 All 289 61 350 289.00 61.00 350.00 Cell Contents: Count Expected count Pearson ChiSquare = 13.766, DF = 1, PValue = 0.000 1 b) Pvalue < 0.005 ...
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This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.
 Spring '09
 KAILASHKAPUR

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