Question

Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1 = 4, s1 = .20, n1 = 15, x⎯⎯2 = 4.25, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) d.f. t-calculated p-value .0088 t-critical -2.145 (a-2) Based on the above data choose the correct decision. Reject the null hypothesis Do not reject the null hypothesis (b-1) Comparison of average commute miles for randomly chosen students at two community colleges: x⎯⎯1 = 17, s1 = 5, n1 = 22, x⎯⎯2 = 21, s2 = 7, n2 = 19, α = .05, two-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) d.f. t-calculated p-value t-critical +/- (b-2) Based on the above data choose the correct decision. Do not reject the null hypothesis Reject the null hypothesis (c-1) Comparison of credits at time of graduation for randomly chosen accounting and economics students: x⎯⎯1 = 141, s1 = 2.8, n1 = 12, x⎯⎯2 = 138, s2 = 2.7, n2 = 17, α = .05, right-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) d.f. t-calculated p-value t-critical (c-2) Based on the above data choose the correct decision. Reject the null hypothesis Do not reject the null hypothesis

Answer 1

Sol:

Part a)

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) + (S_{2}^{2} / n2)}$
$t = ( 4.05 - 4.35 ) / \sqrt{ ( 0.2^{2} / 15) + ( 0.3^{2} / 15)}$
t = -3.2225

Test Criteria :-
Reject null hypothesis if t < -t(α, DF)
$DF = ( ( S_{1}^{2} / n1 &plus; S_{2}^{2} / n2))^{2} / ( (S_{1}^{2} / n1)^{2}/n1-1) &plus; (S_{2}^{2} / n2)^{2}/n2-1) )$
$DF = ( ( 0.2^{2} / 15 &plus; 0.3^{2} / 15 ))^{2} / ( ( 0.2^{2} / 15 )^{2}/ 15 -1) &plus; ( 0.3^{2} / 15 )^{2}/ 15 -1) )$
DF = 24

Critical value t(α, DF) = t( 0.025 , 24 ) = 2.064
t < -t(α, DF) = -3.2225 < -2.064
Result :- Reject Null Hypothesis

Decision based on P value

P - value = P ( t > 3.2225 ) = 0.0018
Reject null hypothesis if P value < α level of significance
P - value = 0.0018 < 0.025 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Part b)

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) &plus; (S_{2}^{2} / n2)}$
$t = ( 19 - 25 ) / \sqrt{ ( 5^{2} / 22) &plus; ( 7^{2} / 19)}$
t = -3.1128

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, DF)
$DF = ( ( S_{1}^{2} / n1 &plus; S_{2}^{2} / n2))^{2} / ( (S_{1}^{2} / n1)^{2}/n1-1) &plus; (S_{2}^{2} / n2)^{2}/n2-1) )$
$DF = ( ( 5^{2} / 22 &plus; 7^{2} / 19 ))^{2} / ( ( 5^{2} / 22 )^{2}/ 22 -1) &plus; ( 7^{2} / 19 )^{2}/ 19 -1) )$
DF = 32

Critical value   t(α/2, DF) = t(0.05 /2, 32 ) = 2.037

| t | > t(α/2, DF) = 3.1128 > 2.037
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 3.1128 ) = 0.0039
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0039 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Part c)

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) &plus; (S_{2}^{2} / n2)}$
$t = ( 144 - 143 ) / \sqrt{ ( 2.8^{2} / 12) &plus; ( 2.7^{2} / 17)}$
t = 0.9613

Test Criteria :-
Reject null hypothesis if t > t(α, DF)
$DF = ( ( S_{1}^{2} / n1 &plus; S_{2}^{2} / n2))^{2} / ( (S_{1}^{2} / n1)^{2}/n1-1) &plus; (S_{2}^{2} / n2)^{2}/n2-1) )$
$DF = ( ( 2.8^{2} / 12 &plus; 2.7^{2} / 17 ))^{2} / ( ( 2.8^{2} / 12 )^{2}/ 12 -1) &plus; ( 2.7^{2} / 17 )^{2}/ 17 -1) )$
DF = 23

Critical value   t(α, DF) = t( 0.05 , 23 ) = 1.714

t > t(α, DF) = 0.9613 < 1.714
Result :- Fail to Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 0.9613 ) = 0.1732
Reject null hypothesis if P value < α level of significance
P - value = 0.1732 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- We Accept H0