Question

PLEASE ANSWER A THROUGH C PLEASE

der the partially completed one-way ANOVA summary table below the remaining entries in the table. Sum of Squares Degrees of Freedom Mean Sum of Squares b) How many population means are being tested? c) Using a 0.05, what conclusions can be made concerning the Within Total 84 164 17 Click the icon to view a table of critical F-scores for ? = 0.05. a) C the ANOVA table below Sum of Degrees of Mean Sum of Squares Freedom Squares 84 164 Within Total Type integers or decimals. Round to three decimal places as needed) nter your answer in the edit fields and then click Check Answer

Answer 1

Solution :

(a) To complete the remaining entries in the table :

The informations that we have from the available data are ,

;  

$df_{Between}=5\ \ ;\ \ df_{Total}=17\ \ ;\ \ \alpha =0.05$

Thus , we have to find out the rest of the data :

$df_{Within}=df_{Total}-df_{Between}=17-5=12$

$Between\ Mean\ Sum\ of\ Squares=MSB=\frac{SSB}{df_{Between}}=\frac{80}{5}=16$

$Within\ Mean\ Sum\ of\ Squares=MSW=\frac{SSW}{df_{Within}}=\frac{84}{12}=7$

$F-ratio=\frac{MSB}{MSW}=\frac{16}{7}=2.286$

$Critical\ Value=F_{crit}=F_{\alpha,df_{Between},df_{Within}}=F_{0.05,5,12}=3.106$

Thus , the complete ANOVA table is given as ,

Source	Sum of Squares	dfs	Mean Sum of Squares	F	Fcrit
Between	SSB = 80	5	MSB = 16	F-ratio = 2.286	3.106
Within	SSW = 84	12	MSW = 7	-	-
TOTAL	SST = 164	17	-	-	-

(b) Here , 6 population means are being tested .

(c) Conclusion : Using $\alpha$ = 0.05 , we can clearly see from the ANOVA table that ,

$2.286<3.106\ \Rightarrow\ F-ratio<F_{crit}$ ; Thus , we accept the null hypothesis of the ANOVA test that all the means are equal at 5% level of significance , and conclude on the basis of the given data that the 6 population means are equal, i.e., the 6 population means do not differ significantly.