Question

1. We would like to see if there is a relationship between heart rates and the number of hours a student studied (0.05 level of significance).  Below is a set of data that was accumulated.  

Hours studied, x	11	10	15	10	6	12	9	10
Blood pressure, y	129	130	130	134	129	131	127	128

Trying to improve the techniques I use in the classroom and reduce the anxiety of the students, I decided to try an experiment.  I decided that I would incorporate more relaxation techniques prior to exams.  For a comparison, I subjected the students to twice the exams (evil, I know).  In essence, I did a pre- and post-test.  The scores are below.  At α = 0.05, can it be concluded that the relaxation techniques helped improve test scores?

No relax	85	72	91	56	80	94	82	78	68
Relax	87	70	92	68	79	93	86	72	70

Answer 1

1)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
11	129	0.39	0.56	-0.47
10	130	0.14	0.06	-0.09
15	130	21.39	0.06	1.16
10	134	0.14	18.06	-1.59
6	129	19.14	0.56	3.28
12	131	2.64	1.56	2.03
9	127	1.89	7.56250	3.7813
10	128	0.14	3.06250	0.656

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	83.00	1038.00	45.88	31.50	8.8
mean	10.38	129.75	SSxx	SSyy	SSxy

sample size ,   n =   8          
here, x̅ = Σx / n=   10.375   ,     ȳ = Σy/n =   129.750  
                  
SSxx =    Σ(x-x̅)² =    45.8750          
SSxy=   Σ(x-x̅)(y-ȳ) =   8.8          
                  
estimated slope , ß1 = SSxy/SSxx =   8.8   /   45.875   =   0.19074
                  
intercept,   ß0 = y̅-ß1* x̄ =   127.77112          
                  
so, regression line is   Ŷ =   127.771   +   0.191   *x

Ho:   ß1=   0          
H1:   ß1╪   0          
n=   8              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    2.230   /√   45.88   =   0.3292
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.1907   /   0.3292   =   0.5794
                  
t-critical value=    2.4469   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   6              
p-value =    0.583417              
decison :    p-value>α , do not reject Ho              
Conclusion:   do not Reject Ho and conclude that linear relations does not exists between X and y
=================

2)

let µd = µpre - µ post

Ho :   µd=   0
Ha :   µd <   0
      

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
85	87	-2.000	0.605
72	70	2.000	10.383
91	92	-1.000	0.049
56	68	-12.000	116.160
80	79	1.000	4.938
94	93	1.000	4.938
82	86	-4.000	7.716
78	72	6.000	52.160
68	70	-2.000	0.605

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	706	717.00	-11.000	197.556

mean of difference ,    D̅ =ΣDi / n =   -1.222                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    4.9694                  
                          
std error , SE = Sd / √n =    4.9694   / √   9   =   1.6565      
                          
t-statistic = (D̅ - µd)/SE = (   -1.222222222   -   0   ) /    1.6565   =   -0.7379
                          
Degree of freedom, DF=   n - 1 =    8                  
  
p-value =        0.2408   [excel function: =t.dist(t-stat,df) ]              
Decision:   p-value>α , Do not reject null hypothesis                      
conclusion:  it cannot be concluded that the relaxation techniques helped improve test scores