Statistics 106
Winter 2020
Final Exam
Due date: March 19 (Thursday), 10am
The exam must be returned online on Canvas, in the form of a .pdf or a .Rmd file.
“The answers here reported constitute my own work. I have consulted the following resources regarding this assignment:” (ADD: names
of persons or web resources, if any, excluding the instructor, TAs, and materials posted on Canvas.)
A team of researchers is testing the effectiveness of the exposure to a specific agent onto the reproductive capability of a type of bacteria. Three levels of exposure to the agent (Low/Moderate/High) are considered. Ten colonies of bacteria are randomly assigned to each level of exposure. The doubling time in hours observed for each colony of bacteria after one month is recorded.
Honor Code:
Colonies of bacteria Total
C1C2C3C4C5C6C7C8C9C10 (Yi·)
ni
nT =
Mean
(Yi·)
Low Moderate High
1.8 2.2 3.3
2.5 2.9 5.3
3.5 4.5 4.8
4.2 4.8 5.7
3.2 5.2 5.2
2.8 3.4 6.7
2.7 4.5 7.3
4.5 4.8 4.5
3.2 3.9 4.8
2.2 3.7 5.7
Y·· =
Y·· =
Exposure to agent
Total
Table 1: Data summary: doubling time for colonies of bacteria with respect to the levels of exposure to a specific agent.
1. Identify factors, their type and nature, treatments, experimental units and the response variable. (3p)
2. Identify the individual sample size n, the number of treatments r, the overall sample size nT . (2p)
3. Which kind of experimental design is this? Is it balanced? (2p)
4. Complete Table 1. Calculate the treatment sum of squares (SSTR), the error sum of squares (SSE), the total sum of squares (SSTO). Then complete Table 2. (5p)
1
Source of Variation Between treatments Within treatments Total
Sum of Squares (SS) SSTR =
SSE = SSTO =
Degrees of Freedom (df)
MS
= SS df
MSTR = MSE =
Table 2: Doubling time for colonies of bacteria with respect to the exposure to a specific agent: analysis of variance.
5. Test the equality of the effects of the different exposure levels onto the doubling time at a significance level of 5%. Comment. (3p)
6. Assume σ = 0.8 and test the null hypothesis μ1· = μ2· at a significance level of 5%. (3p)
7. Suppose that σ = 0.8 and s = ri=1(μi − μ.)2 = 1.2.
Calculate the power of the test used at point 5). (3p)
8. Suppose that σ = 0.8 and the maximum difference among the factor level means is ∆ = 1.2. Which is the minimum sample size required to achieve a power at least equal to δ = 0.8 at a significance level α = 0.05 for the test used at point 5)? (3p)
9. Suppose that σ = 0.8. Test the null hypothesis H0 : μ2 = 3 at a significance level of 5%, against the alternative H1 : μ2 ̸= 3, computing the appropriate test statistics. (3p)
10. In absence of any hypothesis on σ2, Test the null hypothesis H0 : μ2 = 3 at a signifi- cance level of 5%, against the alternative H1 : μ2 ̸= 3, computing the appropriate test statistics. (3p)
11. In absence of any hypothesis on σ2, test the null hypothesis H0 : μ1 = μ2 at a signifi- cance level of 1%, against the alternative H1 : μ1 ̸= μ2, computing the appropriate test statistics. (3p)
12. Derive a confidence interval at a level of 99% for the contrast −1(μ1 + μ2) + μ3. (3p) 2
13. Produce a conditional box-plot of the doubling time with respect to exposure levels. Comment. (3p)
2
14. Produce a normal probability plot of combined studentized ANOVA residuals. Com- ment. (3p)
15. Produce a fitted values VS residuals plot using combined studentized residuals. Com- ment. (3p)
16. Perform Brown-Forsythe test of equal variances at a significance level of 5%. Com- ment. (3p)
Suppose that the researchers want to include in the study the temperature as an ex- plaining factor. In order to do so, they consider three levels of temperature (Low, Medium and High) and they randomly assign ten colonies of bacteria to each combi- nation of exposure and temperature. The doubling time in hours observed for each combination is reported in Table 3.
Temperature (Factor B)
Low Medium High Y i··
Low Moderate High
1.2 2.5 3.2
1.8 3.2 3.4
3.5 4 5
Y·j·
Exposure
(Factor A)
Y··· =
Table 3: Sample means of the doubling time across the levels of exposure and temperature.
17. Identify type, nature and treatments for the new factor. (2p)
18. Which experimental design is this? Is it still balanced? (2p)
19. Identify the individual sample size n, the number of levels of exposure and tempera- ture, a and b, the number of treatments r, the overall sample size nT . (3p)
20. Produce the interaction plot between factor A (exposure) and factor B (temperature), and viceversa, commenting about the main effects of both factors and their interac- tions. (5p)
21. Complete Table 3. Then, calculate the sum of treatment squares for factor A (SSA), factor B (SSB) and their interactions (SSAB). Finally, complete Table 4. (5p)
3
Source of Variation Factor A Factor B Interaction AB Between treatments Within treatments Total
Sum of Squares (SS) SSA =
SSB = SSAB = SSTR = SSE = 200
Degrees of Freedom (df)
MS
= SS df
MSA =
MSB = MSAB = MSTR = MSE =
SSTO =
Table 4: Doubling time by exposure and temperature: two-factor analysis of variance.
22. Test for the overall presence of main effects for factors A and B and of their interac- tions. (4p)
23. Can the row-wise mean μ3· be assumed equal to 3? Perform the individual-wise test and the family-wise test (over the family of row-wise means) at a significance level of 5% and comment. (3p)
24. Test the null hypothesis μ32 − μ31 = 0 at a significance level of 5%, both individual- wise and family-wise (over all pairwise differences involving the simultaneous effects). Comment. (3p)
25. Derive the confidence interval at 95% for the contrast 1 μ.1 + 1 μ.2 −μ.3, both individual- 22
wise and family-wise (over the infinite-dimensional family of contrasts involving all column-wise effects). Comment. (3p)
Suppose now that Table 3 is referred to only one colony of bacteria for each exposure- temperature combination instead of ten.
26. Produce a new ANOVA table with the respective degrees of freedom. (2p)
27. Test for the overall presence of main effects for factors A and B. Are these tests reli- able? (3p)
4
28. Can the row-wise mean μ3· be assumed equal to 3? Perform the individual-wise test and the family-wise test (over the family of row-wise means) at a significance level of 5% and comment. (3p)
29. Estimate the treatment mean for colonies of bacteria with a High exposure and High temperature, and its standard error. Comment. (4p)
30. Perform Tukey’s additivity test with a significance level of 5%. Comment. (3p)
Suppose that the researchers drop the temperature from the study and add a new factor, ’size of bacteria’, as a blocking factor. The size may be Small, Medium, or Large. They now have nine colonies of bacteria, randomly assigned to each temperature-size combination. The resulting ANOVA table is reported in Table 5.
Source of Variation Blocks Treatments Error To t a l
SS SSBL=13.10
SSTR = 26.05 SSBL.TR = 12.89 S S T O = 5 2 . 0 4
df
nb −1=
r − 1 = 2
(nb − 1)(r − 1) = n b r − 1 =
MS MSBL=
MSTR = 13.02 MSBL.TR =
Table 5: ANOVA table for the doubling time wrt to the size of bacteria as a blocking factor.
31. Which experimental design is this? Identify the number of blocks nb, and complete
Table 5. (3p)
32. Test the equality of the effects of the different exposure levels onto the doubling time at a significance level of 5%. Comment. (4p)
5