Question: EXERCISES FOR SECTION 9.1 1. Calculate the tvalue for the following hypotheses, given the sample outcomes indicated. a. H0: ,u =11.H1: ,u #11,} =12.2,standard deviation

 EXERCISES FOR SECTION 9.1 1. Calculate the tvalue for the followinghypotheses, given the sample outcomes indicated. a. H0: ,u =11.H1: ,u #11,}=12.2,standard deviation = 8.9,n at 40. b. HO: ,u =11.H1: y #11,?=14.0,standard deviation = 8.9.n = 40. c. H0: ,u =11.H1: ,u #11,}= 13.7. standard deviation = 8.9,n = 40. d. H0: ,u =234.H1: ,u ;E 234.} = 228.4. standard deviation = 4053. n =120. 2. HO: ,u = 234. Hi: ,u at 234, E =

226.4, standard deviation = 40.53. n = 120. f. HO: ,u =234.H1: ,u at 234.} = 226.4. standard deviation = 40.53. H =120. 2. Determine the probability of t in each example in Exercise1, and decide for each whether you would accept or reject H0.3. Use the Hospital charges worksheet in Ch pt 4-1 .xls. Assumethat the 100 records represent a random samplefrom all hospital discharges fora year and testthe hypothesis H0: Charges = $6, 000 against the

EXERCISES FOR SECTION 9.1 1. Calculate the tvalue for the following hypotheses, given the sample outcomes indicated. a. H0: ,u =11.H1: ,u #11,} =12.2,standard deviation = 8.9,n at 40. b. HO: ,u =11.H1: y #11,? =14.0,standard deviation = 8.9.n = 40. c. H0: ,u =11.H1: ,u #11,} = 13.7. standard deviation = 8.9,n = 40. d. H0: ,u = 234.H1: ,u ;E 234.} = 228.4. standard deviation = 4053. n = 120. 2. HO: ,u = 234. Hi: ,u at 234, E = 226.4, standard deviation = 40.53. n = 120. f. HO: ,u = 234.H1: ,u at 234.} = 226.4. standard deviation = 40.53. H = 120. 2. Determine the probability of t in each example in Exercise 1, and decide for each whether you would accept or reject H0. 3. Use the Hospital charges worksheet in Ch pt 4-1 .xls. Assume that the 100 records represent a random samplefrom all hospital discharges for a year and testthe hypothesis H0: Charges = $6, 000 against the alternative Hi: Charges 36 $6. 000, and indicate your conclusion. EXERCISES FOR SECTION 9.2 1. Assume the following information for equal-sized samples, and make the appropriate t tests. a. Group 1: x = 25, s2 = 27.8, n = 15; Group 2: x = 23, $2 = 27.8, n = 15. b. Group 1: x = 25, s2 = 27.8, n = 15; Group 2:x = 29, $2 = 27.8, n = 15. c. Group 1: X = 125, $2 = 92.6, n = 40; Group 2:x = 121, s2 = 92.6, n = 40. d. Group 1: x = 125, $2 = 92.6, n = 40; Group 2: X = 131, s2 = 92.6, n = 40. e. Group 1: x = 11, $2 = 15, n = 130; Group 2: x = 11.5, s2 = 15, n = 130. f. Group 1:X = 11, $2 = 15, n = 130; Group 2:X = 12.5, $2 = 15, n = 130. 2. Determine the probability of the t test result in (a) through (f) of Exercise 1, and decide whether you would accept or reject the hypothesis of no difference between the two groups in each case.5. Use the data on the Cost worksheet of Chpt 9-1.xis that show the hospital cost for a randomly selected sample of 34 women and 34 men, and do the following: a. Calculate a t test of the hypothesis that cost for women is equal to cost for men against the alternative that they are not equal, assuming equal variance. b. Decide whether to accept or reject the hypothesis of equality.11. Use the Excel add-in for the equal variance t test, and conduct a t test for the data on the Cost worksheet of Chpt 9-1.xIs. Do the results replicate those found in Exercise 5?3. A health payer organization is concerned about the time it takes a group of account assessors to process certain aspects of health care claims. They initiate a training program to reduce the time involved. Average processing time for a xed set of claims is determined before and after the training program. The data are shown on the Time spreadsheet in Chpt 9-2.x|s. 3. Using the formula in Equation 9.1 3. test the hypothesis that processing time before and after training is the same against the alternative that processing time after is less than before, and indicate your conclusion. In. Use the Excel addin for related data, and replicate the results found in (a). The test of the hypothesis, H0, just given, is carried out using the formula in Equation 9'13. dxb-ta t = ' 9.l3 5.5.; { ) where Emma is the mean difference between the before and after measures, SEE is the standard error of the differences, and the degrees of freedom are n 1 where n is the number of difference scores

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