The following data are from a two-factor study examining the effects of two treatment conditions on males and females.
a. Use an ANOVA with α = .05 for all tests to evaluate the significance of the main effects and the interaction.
b. Compute η2 to measure the size of the effect for each main effect and the interaction.
Factor B: Treatment
B1 | B2 | B3 | ||
Male | 3 | 1 | 10 | |
1 | 4 | 10 | ||
1 | 8 | 14 | ||
6 | 6 | 7 | TROW1 =90 ROWI |
|
4 | 6 | 9 | ||
M= 3 | M = 5 | M = 10 | ||
T = 15 | T=25 | T= 50 | ||
SS = 18 | SS =28 | SS = 26 | N = 30 | |
Factor A: | 0 | 2 | 1 | G = 120 |
Gender | 2 | 7 | I | ΣX2 = 860 |
0 | 2 | 1 | ||
0 | 2 | 6 | TROW2 = 30 | |
Female | 3 | 2 | 1 | |
M = 1 | M = 3 | M = 2 | ||
T =5 | T =15 | T =10 | ||
SS =8 | SS =20 | SS = 20 |
TCO4.1 = 20 TCO4.2 = 40 TCO4.3 = 60
a.

Answer to Problem 25P
For factor A treatment has significant effect means there is difference in performance of male and female. For factor B treatment has significant effect means there is difference in treatment with
Explanation of Solution
Given info:
The following data are given in the question:
Factor B | |||||
|
|
|
|||
Factor A | Male | 3 | 1 | 10 |
|
1 | 4 | 10 | |||
1 | 8 | 14 | |||
6 | 6 | 7 | |||
4 | 6 | 9 | |||
|
|
|
| ||
Female | 0 | 2 | 1 |
|
|
2 | 7 | 1 | |||
0 | 2 | 1 | |||
0 | 2 | 6 | |||
3 | 2 | 1 | |||
|
|
|
|||
Total = 20 | Total = 40 | Total = 60 |
Calculation:
Total variability:
Total sum of squareis calculated as:
Total degree of freedomis calculated as:
Within treatment:
Sum of square within treatment is calculated as:
Degree of freedom within treatmentis calculated as:
Between treatment variability:
Sum of square between treatmentsis calculated as:
Degree of freedom between treatments is calculated as:
For factor A, the row totals are 90 and 30, and each total was obtained by adding nothing.
Degree of freedom for factor Ais calculated as:
For factor B, sum of square is calculated as:
Degree of freedom for factor B is calculated as:
Interaction between factors A and B denoted as
Degree of freedom for interaction between factors is calculated as:
Two factor ANOVA consists of three separate hypothesis test with 3 separate F-ratios.
This value is same for all three F-ratios.
The numerator of the 3 F-ratios factor A, Factor B,
Variances for factor Bis calculated as:
Variance for
Now F statistic for factor A is calculated as:
F statistic for factor B is calculated as:
F statistic for
Decision rule:
If the absolute value of the F-ratio is greater than the table value then there is sufficient evidence to reject the null hypothesis.
F-ratio for factor A has
Thus, the absolute F-ratio forfactor A is 24 which is greater than 4.26. Thus, there is enough evidence to reject that there is no mean difference for factor A. Hence, factor A has a significant effect.
Thus, the absolute F-ratio for factor B is * which is greater than 3.40. Thus, there is enough evidence to reject that there is no mean difference for factor A. Hence, factor B has a significant effect.
Thus, the absolute F-ratio forinteraction effect is 6 which is greater than 3.40. Thus, there is enough evidence to reject that there is no mean difference for factor A. Hence, factor B has a significant effect. For these data, treatment of interaction has significant effect.
b.

Compute: The value of
Answer to Problem 25P
The value of
Explanation of Solution
Given info:
The following data are given in the question:
Factor B | |||||
|
|
|
|||
Factor A | Male | 3 | 1 | 10 |
|
1 | 4 | 10 | |||
1 | 8 | 14 | |||
6 | 6 | 7 | |||
4 | 6 | 9 | |||
|
|
|
| ||
Female | 0 | 2 | 1 |
|
|
2 | 7 | 1 | |||
0 | 2 | 1 | |||
0 | 2 | 6 | |||
3 | 2 | 1 | |||
|
|
|
|||
Total = 20 | Total = 40 | Total = 60 |
Calculation:
The size of the effect for factor A is calculated as:
Now substitute the values calculated in part A then:
The size of the effect for factor B is calculated as:
Now substitute the values calculated in part A then:
The size of the effect for interaction
Now substitute the values calculated in part A then:
Conclusion:
The value of
Want to see more full solutions like this?
Chapter 13 Solutions
Bundle: Cengage Advantage Books: Essentials of Statistics for the Behavioral Sciences, 8th + LMS Integrated for MindTap Psychology, 1 term (6 months) Printed Access Card
- Question 1 The data shown in Table 1 are and R values for 24 samples of size n = 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals recorded (i.e., 34.5 should be 0.50345). Table 1: Bearing Diameter Data Sample Number I R Sample Number I R 1 34.5 3 13 35.4 8 2 34.2 4 14 34.0 6 3 31.6 4 15 37.1 5 4 31.5 4 16 34.9 7 5 35.0 5 17 33.5 4 6 34.1 6 18 31.7 3 7 32.6 4 19 34.0 8 8 33.8 3 20 35.1 9 34.8 7 21 33.7 2 10 33.6 8 22 32.8 1 11 31.9 3 23 33.5 3 12 38.6 9 24 34.2 2 (a) Set up and R charts on this process. Does the process seem to be in statistical control? If necessary, revise the trial control limits. [15 pts] (b) If specifications on this diameter are 0.5030±0.0010, find the percentage of nonconforming bearings pro- duced by this process. Assume that diameter is normally distributed. [10 pts] 1arrow_forward4. (5 pts) Conduct a chi-square contingency test (test of independence) to assess whether there is an association between the behavior of the elderly person (did not stop to talk, did stop to talk) and their likelihood of falling. Below, please state your null and alternative hypotheses, calculate your expected values and write them in the table, compute the test statistic, test the null by comparing your test statistic to the critical value in Table A (p. 713-714) of your textbook and/or estimating the P-value, and provide your conclusions in written form. Make sure to show your work. Did not stop walking to talk Stopped walking to talk Suffered a fall 12 11 Totals 23 Did not suffer a fall | 2 Totals 35 37 14 46 60 Tarrow_forwardQuestion 2 Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples of five parts each are collected, and the compressive strengths (in psi) are shown in Table 2. Table 2: Strength Data for Question 2 Sample Number x1 x2 23 x4 x5 R 1 83.0 2 88.6 78.3 78.8 3 85.7 75.8 84.3 81.2 78.7 75.7 77.0 71.0 84.2 81.0 79.1 7.3 80.2 17.6 75.2 80.4 10.4 4 80.8 74.4 82.5 74.1 75.7 77.5 8.4 5 83.4 78.4 82.6 78.2 78.9 80.3 5.2 File Preview 6 75.3 79.9 87.3 89.7 81.8 82.8 14.5 7 74.5 78.0 80.8 73.4 79.7 77.3 7.4 8 79.2 84.4 81.5 86.0 74.5 81.1 11.4 9 80.5 86.2 76.2 64.1 80.2 81.4 9.9 10 75.7 75.2 71.1 82.1 74.3 75.7 10.9 11 80.0 81.5 78.4 73.8 78.1 78.4 7.7 12 80.6 81.8 79.3 73.8 81.7 79.4 8.0 13 82.7 81.3 79.1 82.0 79.5 80.9 3.6 14 79.2 74.9 78.6 77.7 75.3 77.1 4.3 15 85.5 82.1 82.8 73.4 71.7 79.1 13.8 16 78.8 79.6 80.2 79.1 80.8 79.7 2.0 17 82.1 78.2 18 84.5 76.9 75.5 83.5 81.2 19 79.0 77.8 20 84.5 73.1 78.2 82.1 79.2 81.1 7.6 81.2 84.4 81.6 80.8…arrow_forward
- Name: Lab Time: Quiz 7 & 8 (Take Home) - due Wednesday, Feb. 26 Contingency Analysis (Ch. 9) In lab 5, part 3, you will create a mosaic plot and conducted a chi-square contingency test to evaluate whether elderly patients who did not stop walking to talk (vs. those who did stop) were more likely to suffer a fall in the next six months. I have tabulated the data below. Answer the questions below. Please show your calculations on this or a separate sheet. Did not stop walking to talk Stopped walking to talk Totals Suffered a fall Did not suffer a fall Totals 12 11 23 2 35 37 14 14 46 60 Quiz 7: 1. (2 pts) Compute the odds of falling for each group. Compute the odds ratio for those who did not stop walking vs. those who did stop walking. Interpret your result verbally.arrow_forwardSolve please and thank you!arrow_forward7. In a 2011 article, M. Radelet and G. Pierce reported a logistic prediction equation for the death penalty verdicts in North Carolina. Let Y denote whether a subject convicted of murder received the death penalty (1=yes), for the defendant's race h (h1, black; h = 2, white), victim's race i (i = 1, black; i = 2, white), and number of additional factors j (j = 0, 1, 2). For the model logit[P(Y = 1)] = a + ß₁₂ + By + B²², they reported = -5.26, D â BD = 0, BD = 0.17, BY = 0, BY = 0.91, B = 0, B = 2.02, B = 3.98. (a) Estimate the probability of receiving the death penalty for the group most likely to receive it. [4 pts] (b) If, instead, parameters used constraints 3D = BY = 35 = 0, report the esti- mates. [3 pts] h (c) If, instead, parameters used constraints Σ₁ = Σ₁ BY = Σ; B = 0, report the estimates. [3 pts] Hint the probabilities, odds and odds ratios do not change with constraints.arrow_forward
- Solve please and thank you!arrow_forwardSolve please and thank you!arrow_forwardQuestion 1:We want to evaluate the impact on the monetary economy for a company of two types of strategy (competitive strategy, cooperative strategy) adopted by buyers.Competitive strategy: strategy characterized by firm behavior aimed at obtaining concessions from the buyer.Cooperative strategy: a strategy based on a problem-solving negotiating attitude, with a high level of trust and cooperation.A random sample of 17 buyers took part in a negotiation experiment in which 9 buyers adopted the competitive strategy, and the other 8 the cooperative strategy. The savings obtained for each group of buyers are presented in the pdf that i sent: For this problem, we assume that the samples are random and come from two normal populations of unknown but equal variances.According to the theory, the average saving of buyers adopting a competitive strategy will be lower than that of buyers adopting a cooperative strategy.a) Specify the population identifications and the hypotheses H0 and H1…arrow_forward
- You assume that the annual incomes for certain workers are normal with a mean of $28,500 and a standard deviation of $2,400. What’s the chance that a randomly selected employee makes more than $30,000?What’s the chance that 36 randomly selected employees make more than $30,000, on average?arrow_forwardWhat’s the chance that a fair coin comes up heads more than 60 times when you toss it 100 times?arrow_forwardSuppose that you have a normal population of quiz scores with mean 40 and standard deviation 10. Select a random sample of 40. What’s the chance that the mean of the quiz scores won’t exceed 45?Select one individual from the population. What’s the chance that his/her quiz score won’t exceed 45?arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman





