Question 1
The reason for the difference in the σM1-M2 across the two studies:
There is a chance that the distribution of differences and variables varies from study in group A and that of Group B
Question 2
Calculate the σM1 – M2 σM1 – M2 = 1.4 – 1.22 = 0.18
Question 3
Question 4
Complete the following:
Identify the IV and DV in this study.
IV=> receipt of coupon
DV=> purchase amount
State the null hypothesis and the directional (one-tailed) research hypothesis.
Null hypothesis => there is no difference in the amount purchased whether a customer receive or not receive a coupon i.e. there is no difference in the means of purchase amount from both groups.
Run an Independent-Samples T Test using SPSS file assessment4b.sav and paste your output
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She decides to quiz her students at the beginning of class and again at the end of class.
Repeated measures were used
Question 6
Question 7
Compare t with the one-tailed critical t at the .01 α level. Did the students who studied online score significantly higher on the quiz? Give a decision about the null. one-tailed critical t at the .01 α level = 2.539 reject null since our t value of 4.316 > than critical t of 2.539, we reject the null hypothesis which states that there is no significant difference between the means on the quiz between those who did online research and those who watched a movie.
Therefore, from the higher mean of 19.55, students who studied online scored significantly higher on the quiz.
Question 8
Refer back to your SPSS output from Assessment 4 Question 3. Based on that output, what is the 95 percent confidence interval of the difference?
Equal variances assumed 95% CI (17.989, -1.122)
Equal variances not assumed 95% CI (-18.223, -0.889)
Question 9
Assume the within-group deviation is 108.45 and the between-group deviation is 48.68. Calculate the total deviation.
SST = SSB +