DOC PREVIEW
SKIDMORE PS 217 - PS 217 Exam 2

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Exam 2 - 1 of 3 Exam 2 PS 217, Fall 2009 1. Dr. Ty Pest is a human factors psychologist who is interested in testing the impact of different keyboard layouts on typing speed. To that end, he chooses four different conditions: a normal keyboard with normal (QWERTY) keyboard layout, a normal keyboard with the Dvorak layout of keys (supposedly a more efficient layout), an ergonomic (split) keyboard with normal keyboard layout, and an ergonomic (split) keyboard with the Dvorak layout of keys. His participants are ten secretaries who have at least 5 years of typing experience. The DV is the number of words per minute of typing of complex textual material in 15 minutes. Below are the incomplete analyses of his experiment. Complete the analyses and interpret the results as completely as you can. [15 pts] Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared Observed Powera Keyboard Sphericity Assumed 11787 3 3929 344.4 .000 .975 1.000 Error(Keyboard) Sphericity Assumed 308 27 11.41 Decision: Reject H0, because p < .001. € HSD = 3.8711.4110= 4.13 NQ ND EQ ED NQ --- ND 34.5 --- EQ 3.3 31.2 --- ED 37.2 2.7 33.9 --- There is a significant effect of type of keyboard, F(3.27) = 344.4, MSE = 11.41, p < .001, η2 = .975. Post hoc analyses using Tukey’s HSD indicate that typists were faster on the Normal QWERTY keyboard (M = 73.7) and the Ergonomoic QWERTY keyboard (M = 70.4) compared to both the Normal Dvorak keyboard (M = 39.2) and the Ergonomic Dvorak keyboard (M = 36.5). However, you should note that the study could not have been properly counterbalanced, because to do so with four condition (and using complete counterbalancing) would require 24 orders and, therefore, 24 participants. 2. Dr. Justin Case was interested in determining if temperature had an impact on performance on a quiz. He took a class of 100 statistics students and randomly assigned a quarter of the class to a room where the temperature was 50°, another quarter was assigned to a room where the temperature was 70°, another quarter was assigned to a room where the temperature was 80°, and the final quarter was assigned to a room where the temperature was 100°. The students all took the same quiz (max.score = 10). Using the summary data below, conduct as complete an analysis of this experiment as you can, then provide a detailed interpretation of the results of the experiment. [15 pts] 50° 70° 80° 100° T (ΣX) 175 215 200 185 G = 775 Mean 7.0 8.6 8.0 7.4 ΣX2 = 6331 SS 127 36 67 58Exam 2 - 2 of 3 Source SS df MS F Between 36.75 3 12.25 4.08 Within 288 96 3 Total 324.75 99 FMax = 127/36 = 3.53 FMaxCrit = 3.0, thus concerns about heterogeneity of variance would lead you to use α = .01 (to control for an inflated chance of Type I Error). FCrit(3,96) = 3.99. Decision: Reject H0, because FObt ≥ FCrit. € HSD = 3.7325= 1.28 50° 70° 80° 100° 50° --- 70° 1.6 --- 80° 1.0 .6 --- 100° .4 1.2 .6 --- There was a significant effect of temperature, F(3,96) = 4.08, MSE = 3, p < .01, η2 = .11. Post hoc tests using Tukey’s HSD indicate that people in the 50° condition perform significantly worse (M = 7.0) than people in the 70° condition (M = 8.6). 3. Suppose that in the not-too-distant future you conduct a single factor experiment using a repeated measures design, but you can't remember how to perform a repeated measures ANOVA. All you can remember from your PS217 class is how to conduct an independent groups ANOVA, so you go ahead and analyze the data from the repeated measures design using an independent groups ANOVA. Your analysis is significant (i.e., you can reject H0). Even though you've conducted an inappropriate analysis, would you be inclined to believe the results of your analysis? Why or why not? Under which circumstances might you be disinclined to believe the results of your analysis? [10 pts] You should be inclined to believe that the effects are significant. Because the RM design is more powerful, you would expect that the F would be larger for a RM analysis and smaller for an IG analysis. Thus, if the F is significant with an IG analysis, you should expect that it would also be significant with a RM analysis. However, if the individual differences are minor, then you might find that the results are significant with an IG analysis, but the F wouldn’t be significant with a RM analysis. 4. Suppose that you’ve analyzed a set of data from an experiment as a repeated measures ANOVA. You’ve obtained F(3,12) = 5. Your MSError = 1 and there are 5 scores in each condition. Now, however, you find out that you should have computed an independent groups ANOVA. Fill in the source table below as it should appear with the appropriate analysis (i.e., assuming that all the scores are the same). [Hint: Pay attention to the df. If you’re confused, first try to construct as much of the repeated measures source table as you can.] [5 pts] Source SS df MS F Between 15 3 5 2.5 Within 32 16 2 Total 47 19Exam 2 - 3 of 3 Here’s what the RM ANOVA would have looked like: Source SS df MS F Between 15 3 5 5 Within 32 16 Subj 20 4 Error 12 12 1 Total 47 19 5. A researcher studying the impact of caffeine on life expectancy adds two types of caffeine (i.e., found in coffee and chocolate) to the water supply of groups of laboratory-bred rats. This species typically survives about 13 months. The water supply of a control group of rats did not have any caffeine added. The DV is the number of days that each rat lives. Analyze the results of this experiment as completely as you can. What would you do next? [10 pts] Coffee Caffeine Chocolate Caffeine No Caffeine 398 401 412 379 420 386 423 413 394 389 396 409 418 406 415 383 410 401 387 406 394 390 409 398 ANOVA dayslive Sum of Squares df Mean Square F Sig. Between Groups 554.4 2 277.2 1.99 .162 Within Groups 2927.6 21 139.4 Total 3482 23 No need for Hartley’s FMax, because the results aren’t significant. Decision: Retain H0, because p > .05. You need more power for this study: (1) increase sample size (n), (2) increase the treatment (e.g., using more concentrated caffeine), (3) decrease individual differences (e.g., rats all from the same litter, or genetic background), and (4) decrease random variability (e.g., be sure to feed all rats at the same time, be sure that cages and testing situation are


View Full Document

SKIDMORE PS 217 - PS 217 Exam 2

Download PS 217 Exam 2
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view PS 217 Exam 2 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view PS 217 Exam 2 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?