Questions
Note: To help you check your responses, I’ve supplied some rough answers. You should write the whole answers on the exam.
Question 1
An advertising firm is interested in studying the effect of four types of sales advertisements (A, B, C and D) on the sales amount of a certain product in a city. The stores in the city can be divided into three groups as small scale, middle scale and large scale accordingly to the size of their sales amount. Under each group, a random ordering of four types of sales advertisements was exhibited in the selected four stores of same size. Sales amounts for a week are shown in the table below.
advertisement_type store sales
1 A Small 24
2 A Middle 40
3 A Large 41
4 B Small 28
5 B Middle 48
6 B Large 52
7 C Small 20
8 C Middle 32
9 C Large 35
10 D Small 26
11 D Middle 50
12 D Large 55Part 1
These data were given to two students separately and asked to construct the ANOVA table to analyze. The results obtained by them are shown below. Some values were missing in each ANOVA table and marked with a blank space “…”.
Complete each ANOVA table by completing the missing values.
Results of the first student
| Source | DF | SS | MS | F | P |
|---|---|---|---|---|---|
| Advertisement type | … | … | … | … | 0.432 |
| Error | … | 1119.33 | … | ||
| Total | … | 1548.91 |
Results of the second student
| Source | DF | SS | MS | F | P |
|---|---|---|---|---|---|
| Advertisement type | … | … | … | … | 0.006 |
| Store type | … | 1048.17 | … | … | 0.000 |
| Error | … | 71.17 | … | ||
| Total | … | 1549.92 |
Part 2
State the differences between the two approaches these students had applied.
Part 3
One student has obtained the following graph. Comment on it.
Part 4
If you are to perform the same analysis, which approach would you apply? Justify your answer.
Part 5
Write the effect model for the selected approach and define all the terms of it.
Part 6
Write the hypotheses of the selected model and state your decisions at 5% level of significance. Interpret your results. (No need to do multiple comparisons.)
Q1-1: Answer (Help)
First student
Analysis of Variance Table
Response: sales
Df Sum Sq Mean Sq F value Pr(>F)
advertisement_type 3 429.58 143.19 1.0234 0.432
Residuals 8 1119.33 139.92 Second student
Analysis of Variance Table
Response: sales
Df Sum Sq Mean Sq F value Pr(>F)
advertisement_type 3 429.58 143.19 12.073 0.005935 **
store 2 1048.17 524.08 44.185 0.000257 ***
Residuals 6 71.17 11.86
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Q1-2: Answer (Help)
First student used CRD.
Second student used RCBD.
Q1-3: Answer (Help)
Your turn, interpret!
Q1-4: Answer (Help)
RCBD
Reason: Your turn, explain!
Q1-5: Answer (Help)
See lecture materials at https://smart-doe.netlify.app/slides/l6_doe#20
Q1-6: Answer (Help)
See lecture materials at https://smart-doe.netlify.app/slides/l6_doe#25
Use effect model format.
Question 2
What is the main advantage of a randomized complete block design over a completely randomized design?
Question 2: Answer (Help)
Refer to teaching materials
Question 3
What are the advantages of factorial design?
Question 3: Answer (Help)
Refer to teaching materials
Question 4
What is a factorial design?
Question 4: Answer (Help)
In a factorial design, multiple independent variables are tested.
Refer to teaching materials for more information.
Question 5
An experiment is conducted to study the influence of material type and thickness in the strength of screen protectors. The following data are collected.
temperature material strength
1 100 1 581
2 100 1 567
3 100 1 572
4 100 2 555
5 100 2 530
6 100 2 580
7 100 3 556
8 100 3 585
9 100 3 600
10 125 1 1090
11 125 1 1087
12 125 1 1085
13 125 2 1069
14 125 2 1034
15 125 2 999
16 125 3 1046
17 125 3 1054
18 125 3 1070
19 150 1 1392
20 150 1 1380
21 150 1 1388
22 150 2 1330
23 150 2 1315
24 150 2 1300
25 150 3 887
26 150 3 905
27 150 3 890The following summary statistics are obtained
# A tibble: 3 × 2
temperature mean
<fct> <dbl>
1 100 570.
2 125 1059.
3 150 1199.# A tibble: 3 × 2
material mean
<fct> <dbl>
1 1 1016.
2 2 968
3 3 844.# A tibble: 9 × 3
# Groups: material [3]
material temperature mean
<fct> <fct> <dbl>
1 1 100 573.
2 1 125 1087.
3 1 150 1387.
4 2 100 555
5 2 125 1034
6 2 150 1315
7 3 100 580.
8 3 125 1057.
9 3 150 894 mean
1 942.4815Part 1
Use \(\alpha=0.05\) in the analysis. Is there a significant interaction effect? Does material type or thickness affect the response? What conclusions can you draw?
Part 2
Draw graph of mean responses at each treatment combination and comment on it.
Question 5 - Part 1 Answer (Help)
Based on the given information you should be able to construct ANOVA table.
Furthermore, you should be able to compute \(y_{..}\), \(y_{.jk}\), etc.
Using the following information you can complete the ANOVA table.
| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| temperature | 2 | 1964718.296 | 982359.1481 | 3036.4851 | 0 |
| material | 2 | 142091.185 | 71045.5926 | 219.6029 | 0 |
| temperature:material | 4 | 288331.926 | 72082.9815 | 222.8094 | 0 |
| Residuals | 18 | 5823.333 | 323.5185 | NA | NA |
Question 5 - Part 2 Answer (Help)
Question 6
MCQ
Why would you use the Tukey multiple comparison?
To test for normality
To test for homogeneity of variance
To test for differences in pairwise means
To test independence of errors
Question 6: Answer
- To test for differences in pairwise means
Question 7
The following Tukey multiple comparisons results are obtained for data in Question 5. Which combination(s) of treatments would you recommend? State all.
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = strength ~ material * temperature, data = df)
$material
diff lwr upr p adj
2-1 -47.77778 -69.4175 -26.13805 6.82e-05
3-1 -172.11111 -193.7508 -150.47139 0.00e+00
3-2 -124.33333 -145.9731 -102.69361 0.00e+00
$temperature
diff lwr upr p adj
125-100 489.7778 468.1381 511.4175 0
150-100 629.0000 607.3603 650.6397 0
150-125 139.2222 117.5825 160.8619 0
$`material:temperature`
diff lwr upr p adj
2:100-1:100 -18.33333 -69.79110 33.124435 0.9339075
3:100-1:100 7.00000 -44.45777 58.457768 0.9998856
1:125-1:100 514.00000 462.54223 565.457768 0.0000000
2:125-1:100 460.66667 409.20890 512.124435 0.0000000
3:125-1:100 483.33333 431.87557 534.791101 0.0000000
1:150-1:100 813.33333 761.87557 864.791101 0.0000000
2:150-1:100 741.66667 690.20890 793.124435 0.0000000
3:150-1:100 320.66667 269.20890 372.124435 0.0000000
3:100-2:100 25.33333 -26.12443 76.791101 0.7253054
1:125-2:100 532.33333 480.87557 583.791101 0.0000000
2:125-2:100 479.00000 427.54223 530.457768 0.0000000
3:125-2:100 501.66667 450.20890 553.124435 0.0000000
1:150-2:100 831.66667 780.20890 883.124435 0.0000000
2:150-2:100 760.00000 708.54223 811.457768 0.0000000
3:150-2:100 339.00000 287.54223 390.457768 0.0000000
1:125-3:100 507.00000 455.54223 558.457768 0.0000000
2:125-3:100 453.66667 402.20890 505.124435 0.0000000
3:125-3:100 476.33333 424.87557 527.791101 0.0000000
1:150-3:100 806.33333 754.87557 857.791101 0.0000000
2:150-3:100 734.66667 683.20890 786.124435 0.0000000
3:150-3:100 313.66667 262.20890 365.124435 0.0000000
2:125-1:125 -53.33333 -104.79110 -1.875565 0.0388843
3:125-1:125 -30.66667 -82.12443 20.791101 0.5092032
1:150-1:125 299.33333 247.87557 350.791101 0.0000000
2:150-1:125 227.66667 176.20890 279.124435 0.0000000
3:150-1:125 -193.33333 -244.79110 -141.875565 0.0000000
3:125-2:125 22.66667 -28.79110 74.124435 0.8213687
1:150-2:125 352.66667 301.20890 404.124435 0.0000000
2:150-2:125 281.00000 229.54223 332.457768 0.0000000
3:150-2:125 -140.00000 -191.45777 -88.542232 0.0000006
1:150-3:125 330.00000 278.54223 381.457768 0.0000000
2:150-3:125 258.33333 206.87557 309.791101 0.0000000
3:150-3:125 -162.66667 -214.12443 -111.208899 0.0000001
2:150-1:150 -71.66667 -123.12443 -20.208899 0.0030058
3:150-1:150 -492.66667 -544.12443 -441.208899 0.0000000
3:150-2:150 -421.00000 -472.45777 -369.542232 0.0000000Question 7: Answer (Help)
Hint: If interaction is significant, you should compare all material::temperature cell means to determine which ones differ significantly.