mont4e_sm_ch09_sec08 - Section 9-8 9-93 1 The variable of...

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Unformatted text preview: Section 9-8 9-93 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. P-value = 0.070 (using Minitab) 9-94 1. The variable of interest is calls by surgical-medical patients. 2. H0:Calls by surgical-medical patients are independent of Medicare status. 3. H1:Calls by surgical-medical 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. P-value = 0.85 9-46 9-95. 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. P-value = 0.002 9-96 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. P-value = 0.652 9-97. 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. P-value = 0.013 9-98 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: 9-47 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. P-value 0 9-99 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 Chi-Square = 13.766, DF = 1, P-Value = 0.000 1 b) P-value < 0.005 ...
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