6)

Consider the partially completedtwo-way ANOVA summary table. Determine the mean square error for this ANOVA procedure.

Source	Sum of Squares	Degrees of Freedom	Mean Sum of Squares	F
Factor B		2
Factor A	600		200
Interaction	144
Error	384	12
Total	1,288	23

A.32B.21C.26D.43

7)A bank would like to determine if a difference exists in the average credit score between residents of threestates, and also investigate if age plays a role in credit score. The credit scores from a random sample of residents from each state was recorded and residents were categorized as being under 40 years old or 40 years and older. Atwo-way ANOVA was conducted using a=0.05 with Factor A assigned the state residence and Factor B assigned the age group. Determine the number of replications for thistwo-way ANOVA procedure. LOADING... Click the icon to view the summarized ANOVA table.

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	14400	1	14400	4.728063	0.037692	4.170877
Columns	24414	2	12207	4.008019	0.028663	3.31583
Interaction	8272.667	2	4136.333	1.358114	0.272502	3.31583
Within	91369.33	30	3045.644
Total	138456	35

A.6B.30C.2D.1

8)A bank would like to determine if a difference exists in the average credit score between residents of threestates, and also investigate if age plays a role in credit score. The credit scores from a random sample of residents from each state was recorded and residents were categorized as being under 40 years old or 40 years and older. Atwo-way ANOVA was conducted using alphaequals=0.05 with Factor A assigned the state residence and Factor B assigned the age group. Determine a conclusion for this ANOVA procedure. LOADING... Click the icon to view the summarized ANOVA table. A.)The interaction test statistic is more than the criticalvalue; therefore, reject the null hypothesis and conclude that Factors A and B interact.B.)The interaction test statistic is more than the criticalvalue; therefore, do not reject the null hypothesis and conclude that Factors A and B interact.C.)The interaction test statistic is less than the criticalvalue; therefore, do not reject the null hypothesis and can conclude that Factors A and B do not interact.D.)The interaction test statistic is less than the criticalvalue; therefore, reject the null hypothesis and can conclude that Factors A and B do not interact.