In: Statistics and Probability
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 13
What is the standard error of the difference between the means of population 2 and population 3, needed to calculate the Tukey-Kramer q-statistic?
QUESTION 14
What is the Tukey-Kramer q-statistic for populations 2 and 3? (Report the absolute value, if you get a negative number, multiply by -1)
QUESTION 15
For the hemoglobin data of the three populations, what is the critical value of the q-statistic required to reject the null hypothesis of the Tukey-Kramer test at a significance level of 0.05?
QUESTION 16
Which populations are have different means according to the results of the Tukey-Kramer test? Select all that apply, you may need to select more than one correct answer to get credit for this question.
Population 1 and Population 2 |
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Population 1 and Population 3 |
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Population 2 and Population 3 |
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None of the above. All populations have the same means. |
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 | 9.129119 | 0.004606 | 3.982298 |
Within Groups | 4.24322 | 11 | 0.385747 | |||
Total | 11.28629 | 13 |
Since p-value of F test=0.004606<0.05 so these populations have different mean hemoglobin concentrations.
QUESTION 13
QUESTION 14
QUESTION 15
Question 16:
Options:
Population 1 and Population 3
Population 2 and Population 3