![Statistics for the Behavioral Sciences](https://www.bartleby.com/isbn_cover_images/9781506386256/9781506386256_largeCoverImage.gif)
Concept explainers
1.
Complete the F table.
Identify whether the decision to retain or reject the null hypothesis for each hypothesis test.
1.
![Check Mark](/static/check-mark.png)
Answer to Problem 22CAP
The completed F table is,
Source of Variation | SS | df | MS | |
Season | 44,204.17 | 1 | 44,204.17 | 5.139 |
Shift | 1,900.00 | 2 | 950.00 | 0.110 |
Season | 233.34 | 2 | 116.67 | 0.014 |
Error | 154,825.00 | 18 | 8,601.39 | |
Total | 201,162.51 | 23 |
The decision is to reject the null hypothesis for Season, the group means for main effect Season (summer, winter) do significantly vary in the population.
The decision is to retain the null hypothesis for shift, the group means for main effect shift (morning, afternoon, evening) do not significantly vary in the population.
The decision is to retainthe null hypothesis for interaction effect; the means of season do not significantly vary by shift or the combinations of these factors.
Explanation of Solution
Calculation:
From the information given that, the study is based on the eating patterns that might contribute toobesity and these measures the average number of calories (permeal) consumed by shift workers (morning, afternoon, night)during two seasons (summer and winter).
Software procedure:
Step by step procedure to obtain the Two-way between-subjects ANOVA using SPSS software is given as,
- Choose Variable view.
- Under the name, enter the names as Season, Shift andCalories.
- Choose Data view, enter the data for as Season, Shift andCalories, respectively.
- Choose Analyze>General Linear Model>Univariate.
- Enter calories in Dependent variable dialog box.
- Send as Season, Shift into Fixed factors.
- Click on Options button.
- Send Season, Shift into Display Means for dialog box.
- Click on Continue button.
- Click OK.
Output using SPSS software is given below:
The completed F table is,
Source of Variation | SS | df | MS | |
Season | 44,204.17 | 1 | 44,204.17 | 5.139 |
Shift | 1,900.00 | 2 | 950.00 | 0.110 |
Season | 233.34 | 2 | 116.67 | 0.014 |
Error | 154,825.00 | 18 | 8,601.39 | |
Total | 201,162.51 | 23 |
Decision rules:
- If the test statistic value is greater than the critical value, then reject the null hypothesis
- If the test statistic value is smaller than the critical value, then retain the null hypothesis
Hypothesis test for main effect factor A (Season):
Let
Null hypothesis:
That is, the group means for main effect Season (summer, winter) do not significantly vary in the population.
Alternative hypothesis:
That is, the group means for main effect Season (summer, winter) do significantly vary in the population.
Critical value:
The considered significance level is
The degrees of freedom for numerator are 1, the degrees of freedom for denominator are 18 from completed F table.
From the Appendix C: Table C.3-Critical values for F distribution:
- Locate the value 1 in degrees of freedom numerator row.
- Locate the value 18 in degrees of freedom denominator row.
- Locate the 0.05 level of significance (value in lightface type) in combined row.
- The intersecting value that corresponds to the (1, 18) with level of significance 0.05 is 4.41.
Thus, the critical value for
Conclusion:
The value of test statistic is 5.139.
The critical value is 4.41.
The test statistic value is greater than the critical value.
The test statistic value falls under critical region.
Hence the null hypothesis is rejected.
The decision is the group means for main effect Season (summer, winter) do significantly vary in the population.
Hypothesis test for main effect factor B (Shift):
Let
Null hypothesis:
That is, the group means for main effect shift (morning, afternoon, evening) do not significantly vary in the population.
Alternative hypothesis:
That is, the group means for main effect shift (morning, afternoon, evening) do significantly vary in the population.
Critical value:
The considered significance level is
The degrees of freedom for numerator are 2, the degrees of freedom for denominator are 18 from completed F table.
From the Appendix C: Table C.3-Critical values for F distribution:
- Locate the value 2 in degrees of freedom numerator row.
- Locate the value 18 in degrees of freedom denominator row.
- Locate the 0.05 level of significance (value in lightface type) in combined row.
- The intersecting value that corresponds to the (2, 18) with level of significance 0.05 is 3.55.
Thus, the critical value for
Conclusion:
The value of test statistic is 0.110.
The critical value is 3.55.
The test statistic value is less than the critical value.
The test statistic value does not falls under critical region.
Hence the null hypothesis is retained.
The decision is the group means for main effect shift (morning, afternoon, evening) do significantly vary in the population.
Hypothesis test for interaction effect of factor A and B:
Let
Null hypothesis:
That is, the means of season do not significantly vary by shift or the combinations of these factors.
Alternative hypothesis:
That is, the means of season do significantly vary by shift or the combinations of these factors.
Critical value:
The considered significance level is
Thus, the critical value for
Conclusion:
The value of test statistic is 0.014.
The critical value is 3.55.
The test statistic value is less than the critical value.
The test statistic value does not falls under critical region.
Hence the null hypothesis is retained.
The decision is the means of season do not significantly vary by shift or the combinations of these factors.
2.
Explain why the tests of post hoc are not necessary.
2.
![Check Mark](/static/check-mark.png)
Answer to Problem 22CAP
The tests of post hoc are not necessary for this test because only the main effect ‘season’ is significant and another main effect ‘shift’ is not significant.
Explanation of Solution
Justification: The post hoc tests are conducted only when the two main effects in the study are significant. The result of the study is that the main effect ‘season’ is significant this effect has only two levels, but another main effect ‘shift’ is not significant and also the interaction effect bewteen both the factors is not significant. This shows that, conducting post hoc tests is not necessary since the main effect ‘shift’ is not significant.
Want to see more full solutions like this?
Chapter 14 Solutions
Statistics for the Behavioral Sciences
- A retail chain is interested in determining whether a digital video point-of-purchase (POP) display would stimulate higher sales for a brand advertised compared to the standard cardboard point-of-purchase display. To test this, a one-shot static group design experiment was conducted over a four-week period in 100 different stores. Fifty stores were randomly assigned to the control treatment (standard display) and the other 50 stores were randomly assigned to the experimental treatment (digital display). Compare the sales of the control group (standard POP) to the experimental group (digital POP). What were the average sales for the standard POP display (control group)? What were the sales for the digital display (experimental group)? What is the (mean) difference in sales between the experimental group and control group? List the null hypothesis being tested. Do you reject or retain the null hypothesis based on the results of the independent t-test? Was the difference between the…arrow_forwardQuestion 4 An article in Quality Progress (May 2011, pp. 42-48) describes the use of factorial experiments to improve a silver powder production process. This product is used in conductive pastes to manufacture a wide variety of products ranging from silicon wafers to elastic membrane switches. Powder density (g/cm²) and surface area (cm/g) are the two critical characteristics of this product. The experiments involved three factors: reaction temperature, ammonium percentage, stirring rate. Each of these factors had two levels, and the design was replicated twice. The design is shown in Table 3. A222222222222233 Stir Rate (RPM) Ammonium (%) Table 3: Silver Powder Experiment from Exercise 13.23 Temperature (°C) Density Surface Area 100 8 14.68 0.40 100 8 15.18 0.43 30 100 8 15.12 0.42 30 100 17.48 0.41 150 7.54 0.69 150 8 6.66 0.67 30 150 8 12.46 0.52 30 150 8 12.62 0.36 100 40 10.95 0.58 100 40 17.68 0.43 30 100 40 12.65 0.57 30 100 40 15.96 0.54 150 40 8.03 0.68 150 40 8.84 0.75 30 150…arrow_forward- + ++ Table 2: Crack Experiment for Exercise 2 A B C D Treatment Combination (1) Replicate I II 7.037 6.376 14.707 15.219 |++++ 1 བྱ॰༤༠སྦྱོ སྦྱོཋཏྟཱུ a b ab 11.635 12.089 17.273 17.815 с ас 10.403 10.151 4.368 4.098 bc abc 9.360 9.253 13.440 12.923 d 8.561 8.951 ad 16.867 17.052 bd 13.876 13.658 abd 19.824 19.639 cd 11.846 12.337 acd 6.125 5.904 bcd 11.190 10.935 abcd 15.653 15.053 Question 3 Continuation of Exercise 2. One of the variables in the experiment described in Exercise 2, heat treatment method (C), is a categorical variable. Assume that the remaining factors are continuous. (a) Write two regression models for predicting crack length, one for each level of the heat treatment method variable. What differences, if any, do you notice in these two equations? (b) Generate appropriate response surface contour plots for the two regression models in part (a). (c) What set of conditions would you recommend for the factors A, B, and D if you use heat treatment method C = +? (d) Repeat…arrow_forward
- Question 2 A nickel-titanium alloy is used to make components for jet turbine aircraft engines. Cracking is a potentially serious problem in the final part because it can lead to nonrecoverable failure. A test is run at the parts producer to determine the effect of four factors on cracks. The four factors are: pouring temperature (A), titanium content (B), heat treatment method (C), amount of grain refiner used (D). Two replicates of a 24 design are run, and the length of crack (in mm x10-2) induced in a sample coupon subjected to a standard test is measured. The data are shown in Table 2. 1 (a) Estimate the factor effects. Which factor effects appear to be large? (b) Conduct an analysis of variance. Do any of the factors affect cracking? Use a = 0.05. (c) Write down a regression model that can be used to predict crack length as a function of the significant main effects and interactions you have identified in part (b). (d) Analyze the residuals from this experiment. (e) Is there an…arrow_forwardA 24-1 design has been used to investigate the effect of four factors on the resistivity of a silicon wafer. The data from this experiment are shown in Table 4. Table 4: Resistivity Experiment for Exercise 5 Run A B с D Resistivity 1 23 2 3 4 5 6 7 8 9 10 11 12 I+I+I+I+Oooo 0 0 ||++TI++o000 33.2 4.6 31.2 9.6 40.6 162.4 39.4 158.6 63.4 62.6 58.7 0 0 60.9 3 (a) Estimate the factor effects. Plot the effect estimates on a normal probability scale. (b) Identify a tentative model for this process. Fit the model and test for curvature. (c) Plot the residuals from the model in part (b) versus the predicted resistivity. Is there any indication on this plot of model inadequacy? (d) Construct a normal probability plot of the residuals. Is there any reason to doubt the validity of the normality assumption?arrow_forwardStem1: 1,4 Stem 2: 2,4,8 Stem3: 2,4 Stem4: 0,1,6,8 Stem5: 0,1,2,3,9 Stem 6: 2,2 What’s the Min,Q1, Med,Q3,Max?arrow_forward
- Are the t-statistics here greater than 1.96? What do you conclude? colgPA= 1.39+0.412 hsGPA (.33) (0.094) Find the P valuearrow_forwardA poll before the elections showed that in a given sample 79% of people vote for candidate C. How many people should be interviewed so that the pollsters can be 99% sure that from 75% to 83% of the population will vote for candidate C? Round your answer to the whole number.arrow_forwardSuppose a random sample of 459 married couples found that 307 had two or more personality preferences in common. In another random sample of 471 married couples, it was found that only 31 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. Find a95% confidence interval for . Round your answer to three decimal places.arrow_forward
- A history teacher interviewed a random sample of 80 students about their preferences in learning activities outside of school and whether they are considering watching a historical movie at the cinema. 69 answered that they would like to go to the cinema. Let p represent the proportion of students who want to watch a historical movie. Determine the maximal margin of error. Use α = 0.05. Round your answer to three decimal places. arrow_forwardA random sample of medical files is used to estimate the proportion p of all people who have blood type B. If you have no preliminary estimate for p, how many medical files should you include in a random sample in order to be 99% sure that the point estimate will be within a distance of 0.07 from p? Round your answer to the next higher whole number.arrow_forwardA clinical study is designed to assess the average length of hospital stay of patients who underwent surgery. A preliminary study of a random sample of 70 surgery patients’ records showed that the standard deviation of the lengths of stay of all surgery patients is 7.5 days. How large should a sample to estimate the desired mean to within 1 day at 95% confidence? Round your answer to the whole number.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
![Text book image](https://www.bartleby.com/isbn_cover_images/9781119256830/9781119256830_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305251809/9781305251809_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305504912/9781305504912_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780134683416/9780134683416_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781319042578/9781319042578_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781319013387/9781319013387_smallCoverImage.gif)