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 9

1. What is the standard error of the difference between the means of population 1 and population 2, needed to calculate the Tukey-Kramer q-statistic?

QUESTION 10

1. What is the Tukey-Kramer q-statistic for populations 1 and 2? (Report the absolute value, if you get a negative number, multiply by -1)

QUESTION 11

1. What is the standard error of the difference between the means of population 1 and population 3, needed to calculate the Tukey-Kramer q-statistic?

QUESTION 12

1. What is the Tukey-Kramer q-statistic for populations 1 and 3? (Report the absolute value, if you get a negative number, multiply by -1)

Answer 1

Ho:Populations haven't different mean hemoglobin concentrations.

V/s

H1:Populations have different mean hemoglobin concentrations.

Here we use one way anova test.

under Ho,

Using Excel data analysis toopack we solve problem

Enter data in to Excel,

1	14.7	15.22	15.28	16.58	15.1
2	15.36	15.91	14.41	14.73	15.09
3	17.12	16.42	16.43	17.33

Excel = > Data => Data Analysis => Anova : Single factor => input range => group by row => Lables in 1st column => output range => ok

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
1	5	76.88	15.376	0.50408
2	5	75.5	15.1	0.3342
3	4	67.3	16.825	0.2207

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	7.386351	2	3.693176	10.11774	0.003214	3.982298
Within Groups	4.01522	11	0.36502
Total	11.40157	13

test statistic = F =10.12

p-value = 0.003

Here p-value = 0.003 < alpha = 0.05 then we reject H0.

Conclude that these populations have different mean hemoglobin concentrations.

Now we carryout Tukey-Krammer Analysis

where ,

q is value from studentized range distribution table at (between df , within df) and alpha=0.05

	means	n	ABS difference	values	Tukey value
1	15.376	5	|1 - 2|	0.276	0.8410
2	15.1	5	|1- 3 |	1.449	0.8920
3	16.825	5	|2- 3|	1.725	0.8410

QUESTION 9

standard error of the difference between the means of population 1 and population 2

SE = 0.27019

QUESTION 10

Tukey-Kramer q-statistic for populations 1 and 2

Tukey-Kramer q-statistic = 3.1127 * SE = 3.1127 * 0.27019

Tukey = 0.8410

Here |1 - 2| = 0.276 < tukey = 0.8410 then not significant difference between 1 and 2.

QUESTION 11

standard error of the difference between the means of population 1 and population 3

SE = 0.2866

QUESTION 12

Tukey-Kramer q-statistic for populations 1 and 3

Tukey-Kramer q-statistic = 3.1127 * SE

= 3.1127 * 0.2866

Tukey = 0.8920

Here |1 - 3| = 1.449 >  tukey = 0.8920 then significant difference between 1 and 3.