Question

Help with the following question that has a few parts! Thank you!

A doctor is interested in the diabetic status of her male and female patients. She is curious if diabetic status (not diabetic, prediabetic, diabetes) is independent of biological sex (male, female). After pulling the data on 220 patients, she summarizes her findings in the table below: Sex/Status Not Diabetic Prediabetic Diabetic Male 58 33 44 Female 49 11 25 She decides to conduct her test at the 1% significance level.

Part 1

Part 2

Part 3

Part 4

Part 5

Part 6

Part 7

Answer 1

Part 1: Ans: Chi square test of Independence

Part 2 : Ans: Sex and diabetic status are independent

Part 3: Ans : Sex and diabetic status are dependent.

Step-2:Observed Frequencies B1 B2 B3 Total A1 58 33 44 135 A2 49 11 25 85 Total 107 44 69 220 + Step-3:Expected Frequencies Steps Step-3:Expected Frequencies B1 B2 B3 Total A1 65.6591 27 42.3409 135 A2 41.3409 17 26.6591 85 Total 107 44 69 220 Step-4: Compute Chi-square (0; - Ej)2 x2 = 2 Ejj (58 - 65.6591)2 (33 - 27)2 65.6591 27 (44 - 42.3409) 42.3409 + (49 - 41.3409)2 (11 - 17)2 41.3409 17 + (25 - 26.6591 26.6591 58.6617 36 2.7526 58.6617 36 + + + + + 2.7526 26.6591 65.6591 27 42.3409 41.3409 17 = 0.8934 + 1.3333 + 0.065 + 1.419 + 2.1176 + 0.1033 = 5.9316 Step-5: Compute the degrees of freedom (df). df = (2 - 1) · (3 - 1) = 2 Step-6:for 2 df, p(x2 2 5.9316) = 0.9485 Since the p-value(0.9485) > a(0.05) (two-tailed test), we can't reject the null hypothesis Ho

Thus it may conclude that the sex and diabetic status are independent.

Part 7:

the conclusion remain same.

Step-3:Expected Frequencies Steps E1,1 (58 - 63.7885) 63.7885 = 0.5253 E 1,2 (33 - 26.2308)2 26.2308 = 1.7469 E1,3 = (2 - 2.9808) 2.9808 = 0.3227 E2,1 (49 - 43.2115) 43.2115 = 0.7754 E2,2 (11 - 17.7692) 17.7692 = 2.5788 E2,3 = (3 - 2.0192) 2.0192 = 0.4764 Total Step-3:Expected Frequencies B1 B2 B3 A1 63.7885 26.2308 2.9808 A2 43.2115 17.7692 2.0192 93 63 Total 107 44 5 156 Step-4: Compute Chi-square (0ij - Ej) x2 = { Eij (58 - 63.7885)2 (33 - 26.2308)2 63.7885 26.2308 + + (2 - 2.9808)2 (49 - 43.2115)2 (11 - 17.7692)2 (3 - 2.0192) 2.9808 43.2115 17.7692 2.0192 + 33.5063 63.7885 + 45.8225 26.2308 0.9619 2.9808 + 33.5063 43.2115 + 45.8225 17.7692 + 0.9619 2.0192 = 0.5253 + 1.7469 + 0.3227 + 0.7754 +2.5788 +0.4764 = 6.4254 Step-5: Compute the degrees of freedom (df). df = (2 - 1) · (3 - 1) = 2 Step-6:for 2 df, p(x2 2 6.4254) = 0.9598 Since the p-value(0.9598) > a(0.05) (two-tailed test), we can't reject the null hypothesis H..