Question

24)____________       An SAT prep course claims to improve the test scores of students. The

                                     table shows the scores for seven students the first two times they took                    

25)_____________     the verbal SAT. Before taking the SAT for the second time, each

                                     student took a course to try to improve his or her verbal SAT scores.  

                                     Test the claim at a = .05. List the a) null hypothesis b) average difference       

                                   between the scores        

         

Student	1	2	3	4	5	6	7
Score on First SAT	308	456	332	443	306	471	432
Score on Second SAT	421	524	400	421	348	589	391

26)______________   In a crash test at five miles per hour, the mean bumper repair cost for 14

                                    midsize cars was $547 with a standard deviation of $85. In a similar test               

                                    of 23 small cars, the mean bumper repair cost was $347 with a standard                      

27)______________ deviation of $185. At a = 0.05, can you conclude that the mean bumper

                                    repair cost is the same for midsize cars and small cars? List the

                                    26) p-value 27) accept or reject.

Answer 1

Ans ) let d = difference of score on First SAT and score on second SAT

a) null hypothesis is H0:ud = 0

b) average difference  between the scores        

average difference

-113

-68

-68

22

-42

-118

41

using excel we have

t-Test: Paired Two Sample for Means
	Variable 1	Variable 2
Mean	392.5714	442
Variance	5431.286	7069.333
Observations	7	7
Pearson Correlation	0.702503
Hypothesized Mean Difference	0
df	6
t Stat	-2.12297
P(T<=t) one-tail	0.038982
t Critical one-tail	1.94318
P(T<=t) two-tail	0.077963
t Critical two-tail	2.446912

p value = 0.077963 which is greater than 0.05 so we conclude that there is no difference in the score in First SAT and second SAT

Ans _) using excel>addin>phstat>two sample test

we have

Separate-Variances t Test for the Difference Between Two Means
(assumes unequal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.05
Population 1 Sample
Sample Size	14
Sample Mean	547
Sample Standard Deviation	85.0000
Population 2 Sample
Sample Size	23
Sample Mean	347
Sample Standard Deviation	185.0000
Intermediate Calculations
Numerator of Degrees of Freedom	4016476.5598
Denominator of Degrees of Freedom	121135.6921
Total Degrees of Freedom	33.1568
Degrees of Freedom	33
Standard Error	44.7673
Difference in Sample Means	200.0000
Separate-Variance t Test Statistic	4.4675
Two-Tail Test
Lower Critical Value	-2.0345
Upper Critical Value	2.0345
p-Value	0.0001
Reject the null hypothesis

26) p-value = 0.0001

27) reject. Ho