Question

12) A study was constructed to investigate hypnotism on success in math class. The values are the grade on a test before and after hypnosis for the same students. Student 1 2 3 4 5 6 Before 66 65 90 100 85 81 After 68 95 88 95 90 100 a. Fill in the row for ?? (you should use Lists on your calculator and not one at a time) b. State the null and alternative hypothesis: H0:_________________; HA:___________________________ Run the appropriate test on your calculator and use it to answer the following c. The test statistic is: ______________________(include if this is a Z or T score) d.The p-value is e. The conclusion is: (either reject or fail to reject H0 based on a significance level of .05) f. Write a sentence that explains what your conclusion means for the data.

Answer 1

Before	after	d= after-before	(d-dbar)^2
66 65 90 100 85 81	68 95 88 95 90 100	2 30 -2 -5 5 19 M: 8.17	38.03 476.69 103.36 173.36 10.03 117.36 S: 918.83

A) null and alternative hypothesis

H0:d =0

H1:d0

B) test statistics

Difference Scores Calculations

Mean: 8.17
μ = 0
S² = SS⁄df = 918.83/(6-1) = 183.77
S²_M = S²/N = 183.77/6 = 30.63
S_M = √S²_M = √30.63 = 5.53

T-value Calculation

t = (M - μ)/S_M = (8.17 - 0)/5.53 = 1.48

P value = 0.200

Since p value greater than level of significance so we fail to reject H0.

So there insufficient evidence to conclude that there difference between means.