Question

When only two treatments are Involved, ANOVA and the Student's ttest (Chapter 11) result In the same concluslons. Also, for computed test statistics,迢= F To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a hlstory course were divded Into two groups, one consisting of 6 students who took the course In the normal lecture format. The other group of 8 students took the course as a distance course format. At the end of the course, each group was examined with a 50-ltem test. The following Is a list of the number correct for each of the two groups. Traditional LectureDistance 35 37 35 32 31 35 45 38 49 31 32 k here Excel Data File a-1. Complete the ANOVA table. (Round your SS, MS, and Fvalues to 2 decimal places and p value to 4 decimal places.) sS df Factors Error Total a-2 Use aa 0.01 level of significance. (Round your answer to 2 decimal places.) b. Using the ttest from Chapter 11, compute t. (Negative amount should be Indicated by a minus sign. Round your answer to 3 declmal places.) C. There Is any difference In the mean test scores. There is the mean scores between lecture and internet-based formats.

Answer 1

lecture	distance
35	45
37	38
35	49
32	31
31	47
35	32
	43
	43

	lecture	distance
count, ni =	6	8
mean , x̅ i =	34.167	41.000
std. dev., si =	2.229	6.698
sample variances, si^2 =	4.967	44.857

                  
grand mean , x̅̅ =    Σni*x̅i/Σni =                 38.07142857
                  
"square of deviation of sample mean
from grand mean"   15.24716553   8.576530612          
                   TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =    91.4829932   68.6122449           160.0952381
SS(within ) = SSW = Σ(n-1)s² =    24.83333333   314           338.8333333
no. of treatment , k =   2              
df between = k-1 =    1              
df within = N-k =   12              
                  
mean square between groups , MSB = SSB/k-1 =    160.0952381              
                  
mean square within groups , MSW = SSW/N-k =    28.23611111              
                  
F statistic = MSB/MSW =    5.669875624              
P value =   0.034690077              
                  
a-1

anova table
	SS	df	MS	F	p-value
Between:	160.10	1	160.10	5.67	0.0347
Within:	338.83	12	28.24
Total:	498.93	13

.

a-2

critical value of f=9.33

excel formula is { =FINV(0.01,1,12)

b)

mean of sample 1,    x̅1=   34.1667
standard deviation of sample 1,   s1 =    2.2286
size of sample 1,    n1=   6
      
mean of sample 2,    x̅2=   41
standard deviation of sample 2,   s2 =    6.697547526
size of sample 2,    n2=   8
      
difference in sample means =    x̅1-x̅2 =    -6.8333
      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.313766189
      
std error , SE =    Sp*√(1/n1+1/n2) =    2.869761734
      
t-statistic =    ((x̅1-x̅2)-µd)/SE =    -2.381

c)

p-value=0.03469

α = 0.01

Conclusion: p-value >α , Do not Reject null hypothesis

do not reject Ho,there is no difference in the mean scores between lecture and internet based formats.