Question

A researcher would like to evaluate the effectiveness of a pain-relief patch designed for lower back pain. Prior to testing the patch, each of n = 4 patients rates the current level of back pain on a scale from 1 to 8. After wearing the patch for 90 minutes, a second pain rating is recorded. The data are as follows.

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                                                Before        After

                                          __________________

                   G = 36                  6          2

                  G² = 784                6          2

                ΣX² = 192                4          4

                                                8          4

                                         ______________________

                                            T = 24   T = 12

                                          SS = 8    SS = 4

Perform an ANOVA and then analyze the data with a t-test. Use the .05 level of significance for both.

Answer 1

grand mean = (24+12)/8 = 4.50

SS within=SS1 + SS2 = 8+4 = 12

SS(between)= SSB = Σn( x̅ - x̅̅)² = 4(6-4.5)²+4*(3-4.5)² = 9.000 + 9.000 = 18
no. of treatment , k =   2
df between = k-1 =    1
N = Σn =   8
df within = N-k =   6
  
mean square between groups , MSB = SSB/k-1 =    18.0000
  
mean square within groups , MSW = SSW/N-k =    2.0000
  
F-stat = MSB/MSW =    9.0000

	SS	df	MS	F	p-value	F-critical
Between:	18.00	1	18.00	9.00	0.0240	5.99
Within:	12.00	6	2.00
Total:	30.00	7

since, p value=0.0240<α=0.05, test is significant

there is significant difference between means

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using paired t test

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
6	2	4.000	1.000
6	2	4.000	1.000
4	4	0.000	9.000
8	4	4.000	1.000

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	24	12	12.000	12.000

mean of difference ,    D̅ =ΣDi / n =   3.000                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.0000                  
                          
std error , SE = Sd / √n =    2.0000   / √   4   =   1.0000      
                          
t-statistic = (D̅ - µd)/SE = (   3   -   0   ) /    1.0000   =   3.0000
                          
Degree of freedom, DF=   n - 1 =    3                  
  
p-value =        0.0577   [excel function: =t.dist.2t(t-stat,df) ]               
Decision:   p-value>α , Do not reject null hypothesis                      
there is no significant difference between means.