One-way analysis of variance Assignment Example | Topics and Well Written Essays - 1000 words

One-way analysis of variance - Assignment causaBasically, the basis of one-way ANOVA is to partition the sum of squ ares indoors and mingled with classes. This method enables effective coincidence of different classes simultaneously assuming the selective information is normally distributed. One way ANOVA is typesetd in three critical steps starting with obtaining squares for all classes of entropy. The degree of freedom, which is the total itemize of independent data that is considered to estimate a parameter, is also determined. Estimating degrees of freedom later on becomes effective in analysing null hypothesis. match to null hypothesis, the mean of classes under consideration is taken to be the same meaning that the variation within and among classes is not significantly different if not identical. This paper applies one-way ANOVA to analyze data for three categories of doctors. To analyse the variance, one-way ANOVA helps to establish the mean of individual groups, kno wn as the discussion mean. Further, the grand mean, which is the mean for the entire data, is also computed. A scatter diagram (data on appendix) No. of days in NHS only (x-axis) Perform a one-way analysis of variance, recording all your interim calculations. treatment mean for the three groups is NHS only-11.25, private practice only-25.33 and both NHS and private practice-21.92. Grand mean= (11.25+25.33+21.92)/3 = 19.5 reckon the treatment effects of the three groups. =11.25-19.5=-8.25 =25.33-19.5=5.83 =21.92-19.5=2.42 The researcher should then compute one-way ANOVA to determine whether the differences in effects are significant. To determine the variance, the following formula is used One-way ANOVA, MS total = MS Total/ (J-1) = (SS Within +SS between)/ (N-1) MS within estimates variability within a group, it is also known as SS residue or SS error. N is Degree of independence (D.F) calculated as N-1, where N is the total number of observation within individual group. MS wi thin= SS within/ D.F (N-1) On the other hand, MS between estimates variability between the groups, it is also known as SS explained since it shows variability explained by group membership. J is Degrees of Freedom (D.F) calculated as J-1, where J is the total number of observations in all groups. MS between= SS between/ D.F (J-1) Ti=135, Tii=304, Tiii=263 (i) (?y) 2 =7022 = 13,689 N 36 (ii) ?Y2= 122++272+12....+372= 19,578 (iii) ?Ti2 = 1352+ 3042+ 2632 = 1,518.75 +7,701.33+5,764.08 = 14,984.16 N 12 12 12 SS Within= 19,578-14,984.16 = 4,593.84 SS Between=14,984.16- 13,689 =1,295.16 SS Total= 19,578- 13,689= 5,889 Therefore MS Total= SS Total/ (N-1) =5,889/36 =163.58 MS Between= SS Between/ (J-1) =1,295.16/2= 647.58 MS Within= SS Within/ (N-1) =4,593.84/ (36-3) =139.2 artificial lake SS D.F Mean Square F Treatment SS Between= 1,295.16 J-1=2 SS Between/(J-1) =647.58 = MS Between MS Within = 4.7 Error SS Within= 4,593.84 N-J=33 SS Within/(N-1) =139.2 Total SS Total= 5,889 N-1=35 SS Tot al/(N-1) =168.26 Step1 Ho= ?= ?= ?, that is, treatments are equally effective Step2 An F statistic is appropriate measure, since the dependent variable is continuous and there are more than one group. Step 3 Since ? = 0.05 and D.F= 2, 33, accept Ho if F2, 33 19.4 Step4 The computed value of F-statistic is 4.7 Step 5 Accept H0. The treatments are equally effective. Explain what your results mean in a way that a non-statistician could understand. As mentioned above, one-way ANOVA seeks to compare two or more classes of data in order to determine if