Chi-Square, Correlation, and Regression - 1925 Words

Chi-Square, Correlation, and Regression (Statistics Project Sample) Content: Lab 2: ANOVA, Chi-Square, Correlation, and RegressionFirst Last NameName of InstitutionLab 2: ANOVA, Chi-Square, Correlation, and RegressionQuestion 1 (a)Categorical =30 (n %) 31-40 (n %) 41- 50 (n %) 50 (n %) Gender 50 62 51.4 44.7 Exp Fatigue 50 62 51.4 44.7 Exp malaise 50 62 51.4 44.7 Continuous =30 (mean sd) 30-40(mean sd) 4-50(mean sd) 50(mean sd) Alk phosphate Mean = 112.5Sd = 56.49322 Mean = 88.9032Sd = 59.98216 Mean= 106.6667Sd = 41.92374 Mean= 114.5291Sd = 43.25137 Albumin Mean = 3.9313Sd = 0.50637 Mean = 3.971Sd = 0.40585 Mean =3.6333Sd = 0.78776 Mean = 3.6412Sd = 0.55384 Protime Mean = 66.5625Sd = 22.65867 Mean = 70.0Sd = 21.00741 Mean = 54.50Sd = 20.51132 Mean = 55.4706Sd = 22.17866 b) ALK Phosphate lab resultAlbumin lab resultc) H0: there is no significant departure from normality for each of the groups.H1: there is significant departure from normality for each of the groupsResultsALK Phosphate lab result sig. (p à ¢ 0.001) for all the age groups; .0 01, .000, .000, .000Conclusion;We reject the null hypothesis. (P à ¢ 0.001). Each of the levels is not normally distributed therefore not meeting the assumption of normality.ResultsAlbumin lab result; overall the sig. (p) is greater than ÃŽ; .017, .000, .139, .280ConclusionWe fail to reject the null hypothesis, (pÃŽ), hence the groups are normally distributed, meeting the assumption of normality.ResultsProtime lab result sig (pÃŽ) for all age groups .096, .007, .013, .028ConclusionWe retain the null hypothesis. The groups are normally distributed, meeting the assumption.d) H0: There is no difference among Alk Phosphate, Albumin and ProtimeH1: There is a difference among Alk Phosphate, Albumin and Protime.ResultsThe total mean of the three groups, Alk Phosphate, Albumin and Protime was 62.5632(s.d =21.944). The average for Alk Phosphate was the highest 107.75 (s.d = 50.8475) Albumin had the least mean of 4.1067 (s.d = 0.66482) and Protime had 66.5625 (s.d = 22.65867). The sig nificant results F (3,248) = 1.240, p =0.296, F (3,274) = 7.388, p = 0.0 and F (3, 170) = 6.149, p= 0.001 for Alk Phosphate, Albumin and Protime respectively. We can see that there is a significant difference in all the three groups as the value of F P in all the cases.ANOVA Sum of Squares df Mean Square F Sig. ALK Phosphate lab result Between Groups 9803.508 3 3267.836 1.240 .296 Within Groups 653467.810 248 2634.951 Total 663271.317 251 Albumin lab result Between Groups 8.767 3 2.922 7.388 .000 Within Groups 108.390 274 .396 Total 117.157 277 Protime lab result Between Groups 8155.576 3 2718.525 6.149 .001 Within Groups 75155.228 170 442.090 Total 83310.805 173 ConclusionSince the value of F P in all the cases we reject null hypothesis as there is a significant mean different across the groups. A post hoc test has to be performed to determine the differences.ii) ÃŽ= 0.05H0: there is no difference in the 3 groupà ¢Ã¢â€š ¬s variance.H1: there is difference in the 3 groupà ¢Ã¢â€š ¬s variance.Result ALK Phosphate lab result; sig. (p) =0.018.Conclusion We reject the null hypothesis pÃŽ, the assumption is not reasonable.Result: Albumin lab result; sig. (p) =0.054,Conclusion: we retain the null hypothesis pÃŽ. The assumption has been met.Result: Protime lab result; sig. (p) =0.921,Conclusion: we retain the null hypothesis pÃŽ. The assumption has been met.Test of Homogeneity of Variances Levene Statistic df1 df2 Sig. ALK Phosphate lab result 3.415 3 248 .018 Albumin lab result 2.579 3 274 .054 Protime lab result .163 3 170 .921 iii) Turkey HSD= (m1-m2)/à ¢Ã… ¡ (msw/n)ALK Phosphate lab result; msw=2634.951, n=252, k=4, dfw=248Q (0.05) =3.66, Q (0.01) =4.45. Judging by 5% confidence,=3031-40 = (107.75-97.7143)/à ¢Ã… ¡ (2634.951/252)=3.1036 not significantly different=3041-50 = (107.75-105.2333)/à ¢Ã… ¡ (2634.951/252)=0.7784 not significantly different=3050 = (107.75-114.1333)/à ¢Ã… ¡ (2634.951/252)=-1.9744 not significantly different31-4041-50= (97.7143-105.2333)/à ¢Ã… ¡ (2634.951/252)=-2.3257 not significantly different31-4050= (97.7143-114.1333)/à ¢Ã… ¡ (2634.951/252)=-5.0785 significantly different41-5050= (105.2333-114.1333)/à ¢Ã… ¡ (2634.951/252)=-2.7528 not significantly differentAlbumin lab result; msw=.396, n=278, k=4, dfw=274Q (0.05) =3.66, Q (0.01) =4.44=3031-40 = (4.1067-3.8565)/ à ¢Ã… ¡ (0.396/278)=6.6292 significantly different=3041-50 = (4.1067-3.6200)/à ¢Ã… ¡ (0.396/278)=12.8954 significantly different=3050 = (4.1067-3.6788)/à ¢Ã… ¡ (0.396/278)=11.3375 significantly different31-4041-50= (3.8565-3.6200)/ à ¢Ã… ¡ (0.396/278)=6.2662 significantly different31-4050= (3.8565-3.6788)/ à ¢Ã… ¡ (0.396/278)=4.7083 significantly different41-5050= (3.6200-3.6788)/ à ¢Ã… ¡ (0.396/278)=-1.5579 not significantly differentProtime lab result; msw=442.090, n=174, k=4, dfw=170Q (0.05) =3.67, Q (0.01) =4.48. From 5% confidence interval=3031-40= (66.5625-69.5938)/à ¢Ã… ¡ (442.090/174)=-1.9017 not significantly different=3041-50 = (66.5625-53.6000)/à ¢Ã… ¡ (442.090/174)=8.1322 significantly different=3050 = (66.5625-56.7895)/à ¢Ã… ¡ (442.090/174)=6.1322 significantly different31-4041-50= (69.5938-53.6000)/à ¢Ã… ¡ (442.090/174)=10.0339 significantly different31-4050= (69.5938-56.7895)/à ¢Ã… ¡ (442.090/174)=8.0395 significantly different41-5050= (53.6000-56.7895)/à ¢Ã… ¡ (442.090/174)=-2.0010 not significantly differentQuestion 2H0: Gender and Age when categorized are independentH1: Gender and age when categorized are not independentResultsChi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 10.120a 3 .018 Likelihood Ratio 10.415 3 .015 Linear-by-Linear Association .161 1 .689 N of Valid Cases 310 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 6.61. X (1) = (10.120, 3), p= 0.018, thus since p0.05 we reject the null hypothesis.ConclusionWe conclude that there is no statistically significance evidence to proof that associatio n between gender and age when categorized are independent.Question 3H0: Experiencing Malaise and age when categorized are independentH1: Experiencing Malaise and age when categorized are not independentResultsChi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 6.621a 3 .085 Likelihood Ratio 6.676 3 .083 Linear-by-Linear Association 2.774 1 .096 N of Valid Cases 308 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 25.35. The Pearson chi-Square X (6.621, 3) p = 0.085. Since p 0.05 we accept the null hypothesis.ConclusionThere is statistically significant evidence that experiencing Malaise and age when categorized are independent since p (0.085) 0.05Question 4H0: Experiencing fatigue and age when categorized are independentH1: Experiencing fatigue and age when categorized are not independentResultsChi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 18.731a 3 .000 Likelihood Ratio 19.494 3 .000 Linear-by-Linear Associati on 15.371 1 .000 N of Valid Cases 308 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 22.44. The Pearson Chi-Square X (18.731, 3) p= 0.0. Since P (0.0) 0.05, we reject the null hypothesis.ConclusionThere is no statistically significance evidence to proof that experiencing fatigue and age when categorized are independent.Question 5H0: Having an abnormal result in the first trimester is related to having an abnormal result in the third trimester.H1: Having an abnormal result in the first trimester is not related to having an abnormal result in the third trimester.Chi-Square Tests Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided) Pearson Chi-Square 7.073a 1 .008