Question

Consider the following measurements of blood hemoglobin concentrations (in g/dL) from three human populations at different geographic locations:

population1 = [ 14.7 , 15.22, 15.28, 16.58, 15.10 ]

population2 = [ 15.66, 15.91, 14.41, 14.73, 15.09]

population3 = [ 17.12, 16.42, 16.43, 17.33]

Perform ANOVA to check if any of these populations have different mean hemoglobin concentrations. (Assume that all the ANOVA requirements such as normality, equal variances and random samples are met.) After you perform ANOVA perform a Tukey-Kramer post-hoc test at a significance level of 0.05 to see which populations actually have different means. As usual, round all answers to two digits after the decimal point. (Make sure you round off to at least three digits any intermediate results in order to obtain the required precision of the final answers.) For any questions, which ask about differences in means or test statistics, which depend on differences in means provide absolute values. In other words if you get a negative value, multiply by -1 to make it positive.

QUESTION 5

1. For the three populations, what is the value of MSgroups in the ANOVA table?

QUESTION 6

1. For the three populations, what is the value of MSerror in the ANOVA table?

QUESTION 7

1. If you run ANOVA on the data for the three populations, what is the F-ratio?  

QUESTION 8

1. If you run ANOVA on the data for the three populations, what is the range of the P-value?

   		Greater than 0.05
   		between 0.05 and 0.025
   		between 0.025 and 0.01
   		less than 0.01

Answer 1

Ho : There is no significant difference in the population's mean hemoglobin concentrations.

H1 : There is significant difference in the population's mean hemoglobin concentrations.

Level of Significance :- (l.o.s.) alpha = 0.05

Decision Criteria :- Reject Ho at 5% l.o.s if F > F crit

Calculations: Using the Single Factor Annova test from the data analysis package in the Excel, we get the following results :

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Population 1	5	76.88	15.376	0.50408
Population 2	5	75.8	15.16	0.3912
Population 3	4	67.3	16.825	0.2207
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	7.043066	2	3.521533	9.129	0.004606	3.982
Within Groups	4.24322	11	0.385747
Total	11.28629	13

Conclusions : Since F > F crit, we reject Ho at 5% l.o.s. and thus conclude that there is significant difference in the population's mean hemoglobin concentrations.

Tukey-Kramer post-hoc test at a significance level of 0.05 to see which populations actually have different means is been carried out in the Excel as follows:

		Difference	n1	n2	SE	q statistic	q table
Population 1	Population 2	0.216	5	5	0.278	0.778	3.82
Population 1	Population 3	1.449	5	4	0.295	4.918	3.82
Population 2	Population 3	1.665	5	4	0.295	5.652	3.82

Here, No. of cases , a =[ No. of groups*( No. of groups -1 ) ] / 2 = 3*2/2 = 3

Difference = Absolute difference of the means of the two selected populations

SE = sqrt [ 0.5 * MSE *(1/n1 + 1/n2) ] where MSE = 0.385747 from the first table.

q statistic = Difference / SE

Decision Criteria : If q statistic > q table at 5% l.o.s. , then we conclude that the respective population means are significantly different. Here, value from q table at 5% l.o.s is obtained for the degrees of freedom of (a , N - a ) where N is the total number of observations.

Thus, the q table for this problem corresponding to the degrees of freedom of ( 3 , 11 ) is 3.82

Conclusion : Since q statistic > q table at 5% l.o.s. for the second and the third case of the 2nd table, we conclude that the means of population 1 and 3 and means of population 2 and 3 differ significantly from each other.

Question 5 : MSgroups = 3.521533 = 3.522 [ from the first table ]

Question 6 : MSerrors = 0.385747 = 0.386   [ from the first table ]

Question 7 : F-ratio = 9.129   [ from the first table ]

Question 8 : p-value = 0.004606    [ from the first table ]

i.e. less than 0.01