Solution
Let X = Posttest score of Experimental Group , Y = Posttest score of Experimental Group
Then, X ~ N(µ_{1}, σ_{1}^{2}) and Y ~ N(µ_{2}, σ_{2}^{2})
Part (a)
To test equality of population standard deviations.
Hypotheses:
Null H_{0}: σ_{1}^{2} = σ^{2}_{2 } Vs Alternative H_{A}: σ_{1}^{2} ≠ σ^{2}_{2}
Test statistic:
F= s_{1}^{2}/s_{2}^{2}where s_{1}and s_{2} are standard deviations based on n1 observations on X and n2 observations on Y.
= 1.2351
[Note: F is defined as above when s_{1}^{2} > s_{2}^{2}. If s_{2}^{2} > s_{1}^{2}, F is defined as F = s_{2}^{2}/s_{1}^{2}]
Calculations
n1 
38 
n2 
23 
s1 
10.18 
s2 
9.16 
Fcal 
1.235107 
Assuming, 

α = 
0.05 
Fcrit 
1.949457 
n1  1 
37 
n2  1 
22 
pvalue 
0.304313 
Distribution, Critical Value and pvalue
Under H_{0}, F~ F_{n1 – 1, n2 – 1} or F_{n2 – 1, n1 – 1} depending on whether s_{1}^{2} > s_{2}^{2} or s_{2}^{2} > s_{1}^{2}
Critical value = upper α% point of F_{n1 – 1, n2 – 1} or F_{n2 – 1, n1 – 1} depending on whether s_{1}^{2} > s_{2}^{2} or s_{2}^{2} > s_{1}^{2}
pvalue = P(F_{n1 – 1, n2 – 1} > F_{(cal)}) or P(F_{n2 – 1, n1 – 1} > F_{(cal)}), depending on whether s_{1}^{2} > s_{2}^{2} or s_{2}^{2} > s_{1}^{2}
Fcrit and and pvalue are found to be as shown in the above table [using Excel Function: Statistical FINV and FDIST respectively]
Decision:
Since pvalue > α. H_{0} is accepted.
Conclusion:
There is not sufficient evidence to suggest that the population variances and hence the population standard deviations are different for the two groups. Answer
Part (b)
To test equality of population means where σ_{1}^{2} = σ_{2}^{2} = σ^{2}, say but σ^{2} is unknown [vide Answer 1].
Hypotheses:
Null: H_{0}: µ1 = µ2 Vs Alternative: H_{A}: µ1 ≠ µ2
Test Statistic:
t = (Xbar  Ybar)/[s√{(1/n_{1}) + (1/n_{2})}] where
s^{2} = {(n_{1} – 1)s_{1}^{2} + (n_{2} – 1)s_{2}^{2}}/(n_{1} + n_{2} – 2);
Xbar and Ybar are sample averages and s_{1},s_{2} are sample standard deviations based on n_{1} observations on X and n_{2} observations on Y respectively.
= 2.427
Calculations
Summary of Excel calculations is given below:
n1 
38 
n2 
23 
Xbar 
33.42 
Ybar 
39.71 
s1 
10.18 
s2 
9.16 
s^2 
96.27664 
s 
9.812066 
tcal 
 2.426504 
Assuming α = 0.05 

tcrit 
2.000995 
pvalue 
0.018319 
Distribution, Significance Level, α Critical Value and pvalue:
Under H_{0}, t ~ t_{n1 + n2  2}. Hence, for level of significance α%, Critical Value = upper (α/2)% point of t_{n1 + n2  2} and pvalue = P(t_{n1 + n2  2} >  tcal ) = 2P(t_{n1 + n2  2} > tcal) if tcal > 0 and 2P(t_{n1 + n2  2} < tcal) if tcal < 0
Using Excel Function: Statistical TINV and TDIST, these are found to be as shown in the above table.
Decision:
Since  tcal  > tcrit, or equivalently since pvalue < α, H_{0} is rejected.
Conclusion:
There is sufficient evidence to suggest that the two groups differ with respect to the means. i.e., we conclude that the population means of posttest scores differ from group to group. Answer
DONE
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