1. (20 points) An educational psychologist was interested in the application of
ID: 2932410 • Letter: 1
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1. (20 points) An educational psychologist was interested in the application of state dependent learning. She taught a particularly challenging topic to her small class of 42 students and used blue paper for the class handouts. When it was time for the exam in class which included that topic, she had the exam copied in two formats: one on all regular paper, and the other on the same blue paper as was used for the handouts. Summary data are presented below. Did the color of the paper affect exam scores? x=21.5 s=2.20 x=22.5 $21.98 Spooled-2.09 a. Identify the appropriate statistical test. correlated groups t-test b. State the null and alternative hypotheses. HO: M1=M2 H1: M1 (does not equal to sign) M2 c. What is the critical value of the test statistic? What is the decision rule for rejecting the null hypothesis? Draw the picture of the rejection region. CV: d. Calculate the observed value of the test statistic and eta2. Write out both the standard error and Reject Ho if t t formulas, and show your work for obtaining the values. e. Write out a formal conclusion about the research project Use the context of the study in your answerExplanation / Answer
Q.1
(a) we will use two sample t - test with equal variances
(b) Null hypothesis: H0: M1 = M2
Alterative Hypothesis: Ha : M1 M2
(3) Critical value of test statitic
tcr= 1.9893 for dF = 42 + 42 - 2 = 82 and alpha = 0.05
Decision Rule :
reject the null if t > tcr
Standard error of the difference = sp * sqrt(1/n1+ 1/n2) = 2.09 * sqrt(1/42 + 1/42) = 0.456
Test statistic
t = (Xblue- Xregular)/ sp * sqrt(1/n1+ 1/n2)
t = (22.5 - 21.5)/ 0.456 = 2.19
so here t > tcritical so we shall reject the null hypothesis and can conclude that marks when blue handout is distributed is higher then the regular handout distributed.
eta2 = (M2 - M1 )/ sp = (22.5 - 21.5)/ 2.09 = 0.48
so we can say that effect size is moderate here. Not so significant.
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