STAT 200 Week 6 Homework | Complete Solution A++ use as a guide only

18. You choose an alpha level of .01 and then analyze your data.

a. What is the probability thxxxxxxxxxxxxxxxat you will make a Type I error given that the null hypothesis is true?

The probability of type I errorxxxxxxxxxxxxxxxxxxxxxxxxxxx is actually alpha given that the null hypothesis is true so it is 0.01.

b. What is the probability that you will make a Type I error given that the null hypothesis is false?

When null hypothesis is false, it is impossible to make a type I error. It means probability that you will make a type I error given that the null hypothesis is false is zero.

7. Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects get- ting better each trial? Test txxxxxxxxxxxxxxhe linear effect of trial for the data.

a

b

c

4

6

7

3

7

8

2

8

5

1

4

7

4

6

9

2

4

2xxxxxxxxxxxxxxxxxxxxxxxx

a. Compute L for eachxxxxxxxxxxxxxxxxxxxxxx subject using the contrast weights -1, 0, and 1. That is, compute (-1)(a) + (0)(b) + (1)(c) for each subject.

L1

-1*4+0*6+1*7=3

L2

-1*3+0*7+1*8=5

L3

-1*2+0*8+1*5=3

L4xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

-1*1+0*4+1*7=6

L5

-1*4+0*6+1*9=5

L6

-1*2+0*4+1*2=0

b. Compute a one-sample t-test on this column (with the L values for each subject) you created.

M=Sample xxxxxxxxxxMean = (3+5+3+6+5+0)/6 = 3.667

Standard error xxxxxxxxxxxxxxoxxxxxxxxxf mean = Sm = 2.160/sqrt(6) = 0.8819

t=(M-mu)/Sm = 3.6x7/0.8819 = 4.158

Using calculator, wxxxxxxxxxxxxxxxxxe find out the probability of two tailed test to be 0.0088

13. You are conducting a study to see if students do better when they study all at once or in intervals. One groupxxxxxxxxxxxxxxxxxxxx of 12 participants took a test after studying for one hour continuously. The other group of 12 participants took a test after studying for three twenty minute sessions. The first group had a mean score of 75 and a variance of 120. The second group had a mean score of 86 and a variance of 100.