This homework covers chapter 11 of your text. It is worth 3% of your grade. There is an SPSS portion. Please put your name ON this document and also as part of the Edoc. There are 50 points in this hw. Chapter 11 28 points this section 1. For each study design below, choose whether it is a one sample T, an independent T, or a repeated measures T. (1 point each) One sample a. A classroom of students get a math training module and their scores are measured before and after. b. One classroom of students gets a math training module, and a different classroom of students doesn't, and their scores are compared. Independent t c. A classroom of students gets one math training module and then takes a math test. Then the same students get a different training module, and retake the same math test, and the scores are compared. Repeated measures d. A classroom of students gets a math training module, and their test scores are compared to existing national test scores One sample 2. What exactly is MD? How do you get it? How does it relate to the null hypothesis Mew? Actually understanding what this is, and reminding yourself of it clearly, will help keep you from getting confused in this chapter (2 points) Md= mean differences (pg. 354) table 11.2 ----- sum of D scores divided by n 3. The following data were obtained from a repeated-measures research study. What is the value of MD for these data? (1 point) Subject 1st 2nd D #1 15 15 0 #2 5 8 3 #3 7 5 2 #4 4 11 7 sum=12 -2 or 2 4. Describe the steps for calculating the above MD (1 point) Md=sumD/n so 12/4 so md=3 (((to get d you subtract to get the \"difference\" in means ))) 5. In a repeated measures design, what raw scores do you use to get the standard error? 6. A repeated-measures study using a sample of n = 28 participants would produce a t statistic with df = ____. (1 point) df=27 7. A researcher uses a repeated-measures study to compare two treatment conditions with a set of 19 scores in each treatment. What would be the value of df for the repeated-measures t statistic? Think about this carefully. (hint: be sure you understand what is n in a repeated measures study, and how you get the df?) Df=18 8. A repeated-measures study and an independent-measures study both produced a t statistic with df = 22. How many individuals participated in each study? (hint: ITS NOT THE SAME NUMBER FOR BOTH!) (1 point) df of rm = 23 df of im= 12 ((you take 22/2 and add 1)) 9. Explain your answer to number 7. (1 point) Df of a repeated measures test (each individual takes part in each treatment) if there are 19 scores then there are 19 participants and n-1 would be 19-1 = 18 so df=18 10. What is the correct null hypothesis for a repeated-measures t test? (1 point) Population mean for each treatment is equal to each other treatment mew1=mew2=mew3 and so on 11. A sample of difference scores has a mean of MD = 5 with a variance of s2 = 144. If effect size is measured using Cohen's d, what is the value of d? (1 point) 5/144=0.035 d=0.035 12. A researcher obtains t(20) = 4.00 and MD = 9 for a repeated-measures study. If the researcher measures effect size using the percentage of variance accounted for, what value will be obtained for 2? (1 point) (4/\\2) / 4/\\2 = 20 n/\\2=0.444 13. A researcher conducts a repeated-measures study to evaluate a treatment with a sample of n = 16 participants and obtains a t statistic of t = 1.94 and a sig (or p) value for that specific T value of .03. The treatment is expected to increase scores and the sample mean shows an increase. Accept or reject the null for a one tailed test using = .05? (1 point) P=0.03 < 0.05 sig. value we reject the null 14. A research report describing the results from a repeated-measures study states: The data show no significant difference between the two treatments, t(10) = 1.65, p > .05. Based on this report, you can conclude that a total of __11__ individuals participated in the research study. (1 point) 15. Compared to an independent-measures design, a repeated-measured study is more likely to find a significant effect because it reduces the contribution of variance due to _individual differences_____. (1 point) 16. A sample of college students is used to test the effectiveness of a new Study Skills Workshop. Each student's grade point average (GPA) is recorded for the semester before the workshop and for the semester after the workshop. The average GPA improved by MD = 0.60. The 95% confidence interval is .60 plus or minus .15. What does that mean in terms of the actual numerical improvement in gpa? (1 point) 17. Briefly explain the advantages and disadvantages of using a repeated-measures design as opposed to an independent-measures design and given an example (2 points). Don't just say whatever comes into your head, have a look in your text. your results are less likely to be skewed or distorted by individual differences when you use a repeated-measures design since your participant is being compared only against themselves as opposed to others, plus having one person try multiple conditions saves time and resources, having too many participants can get out of hand and hard to control for. 18. A researcher wants to examine how the chemical tryptophan, contained in foods such as turkey, can affect mental alertness, at an alpha of .05, two tailed. A sample of n = 9 college students is obtained and each student's performance on a familiar video game is measured before and after eating a traditional Thanksgiving dinner including roast turkey. The average decrease in scores was M = 12 points after the meal with variance = 144 for the difference scores. Answer the questions below (6 points total this problem). a. What are the df? (1 pt) Df=8 b. What is the Standard error? (1 point) 12/3=4 so SE= 4.00 c. What is the calculated value of T? (1 point) -12/4=3 T=-3 d. Assume the critical cutoff for a two tailed T is plus or minus 2.306. Is your finding significant or not? Say something about why you can or cannot reject the null. (1 point) Since t=-3 which is smaller than -2.306 (critical value) ((cv being +/- 2.306)) then we would say that what we found was significant and would reject the null hypothesis e. Compute 2 to measure the size of the effect. (1 point) -3/\\2 divided by -3/\\2 +8 =0.529 or 0.53 f. Write a sentence demonstrating how the outcome of the test and the measure of effect size would appear in a research report. (1 point) Since our outcome showed significance, there is enough evidence to say that there is a significant difference in an average student's video game performance when compared to before and after eating Thanksgiving dinner. SPSS Portion Here you analyze T tests in SPSS, and interpret output. 22 points total this section Part I Independent Sample (output 2 points, questions 9 points) Enter the following data in SPSS. Be sure the data are aligned as shown, since you'll be running t tests. The first column is weight. The second is animal typecat or dog. 1=cat and 2 = dog. You will run a t test to compare whether dogs or cats weigh more. Wt 2.00 2.00 3.00 2.00 4.00 2.00 3.00 2.00 3.00 2.00 3.00 2.00 3.00 5.00 5.00 5.00 7.00 7.00 7.00 6.00 Type 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 2.00 In SPSS go to Analyze: Compare Means: Independent Samples. You will see a dialogue box that says test variables and grouping variables. Click on the weight variable to go in the test variable box, and the type variable to go in the grouping variable box. Under the grouping variable box you will see where it says \"define groups?\" Click on that. Another dialogue box will open which says \"user specified values.\" Put a 1 in group 1 and a 2 in group 2 (you have just told the program that your data is divided into two groups, and that the groups are identified with the number 1 and the number 2.) Then hit OK.,. An output box will open. Copy and paste that output into your hw doc. Use the output to respond to the following questions: 1) Looking at the group statistics box, what is the mean and standard deviation for weight in group one and group two? (2 points) 2) Look at the next box that says Independent Samples Test. Ignore the first few columns about Levene's equality of variancesthis is not the outcome of the t test, you don't need to focus on that right now. Go right to the INDEPENDENT SAMPLES t TEST COLUMN. a. What are the t and the df in the first row of data in this box? (2 pts) b. What is the significance level? What does this mean about errors we might commit? (2 pts) 0.001 or 1% probability that the average weight is equal between does and cats (not likely) c. Are the weights for cats and dogs different or not different, according to this test? (1 pt) Probability is very very small, so we reject the null since it is likely that they are different weights (means are different) between the species d. What is the 95% confidence interval for the difference, and what do these numbers mean in words? (2 pts) -3.80, -1.20 ---we are 95 % sure that the difference between mean weights for each species (dog vs. cat) is somewhere falling between -3.80 and -1.20, the interval doesn't have a zero in it we say that dogs on average weigh more than cats do Part II Repeated measures 2 points output, 9 points questions. Now, enter the following data, again, being sure to keep it aligned. We will pretend this is weight data for kittens on a weight gaining diet. The first column is their weight before, and the second column is the same kittens weighed after. 2.00 2.00 3.00 2.00 4.00 2.00 3.00 2.00 3.00 2.00 3.00 2.00 3.00 5.00 5.00 5.00 7.00 7.00 7.00 6.00 Once you have entered the data, go to analyze, compare means, paired samples. A dialogue box will open, with a spot for \"paired variables.\" Put one variable into one box, and the other right next to it (not below it)., Then click \"ok.\" You will get more output. Paste it into your hw doc. 1. Looking at the group statistics box, a. what is the mean and standard deviation for their weight at time one at time two? (1 point) Time one: mean is 2.50 and SD= 0.7071 Time two: mean 5.00 and SD= 1.8257 b. Does this look familiar? Where have you seen this before? (1 point) Looks like the above problem about animals (cats vs. dogs) it's just like weight 2. Now look below at the paired samples test. a. What is the t value? (1 point) T=-4.294 b. Is it significant? (1 point) Probability is .002 so yes it's significant c. Is it the same or different from the T value from the test above? (1 point) -4.294 is not the same as -4.0338 from above that d. If the T value is different, why do you think that is? Really think about this. This is asking you to consider what the real difference between an independent samples and related samples test is. Think about what you have learned about individual differences leading to more variation, and the meaning of the standard error. (4 points) The SE in the paired test was higher than the independent test where it was lower, so the t=higher for paired and t=lower for independent test, the same participants were used across the board for the paired test which affected its outcome