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¯1x¯1 = 4, s₁ = .20, n₁ = 15, x¯2x¯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
t-critical

(a-2) Based on the above data choose the correct decision.
  

• Do not reject the null hypothesis

• Reject the null hypothesis

(b-1) Comparison of average commute miles for randomly chosen students at two community colleges:

x¯1x¯1 = 17, s₁ = 5, n₁ = 22, x¯2x¯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¯1x¯1 = 141, s₁ = 2.8, n₁ = 12, x¯2x¯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.
  

• Do not reject the null hypothesis

• Reject the null hypothesis

Answer 1

Part a-1)

DF = 24

Test Statistic :-


t = -2.6854

P - value = P ( t > 2.6854 ) = 0.0065

Critical value t(α, DF) = t( 0.025 , 24 ) = 2.0639

Part a-2)

Test Criteria :-
Reject null hypothesis if t < -t(α, DF)
t < -t(α, DF) = -2.6854 < -2.064
Result :- Reject Null Hypothesis

Decision based on P value
Reject null hypothesis if P value < α level of significance
P - value = 0.0065 < 0.025 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Part b-1)

DF = 32

Test Statistic :-


t = -2.0752

P - value = P ( t > 2.0752 ) = 0.0461

Critical value t(α/2, DF) = t(0.05 /2, 32 ) = ± 2.0369

Part b-2)
Test Criteria :-
Reject null hypothesis if | t | > t(α/2, DF)
| t | > t(α/2, DF) = 2.0752 > 2.037
Result :- Reject Null Hypothesis

Decision based on P value
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0461 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Part c-1)

DF = 23

Test Statistic :-


t = 2.8839

P - value = P ( t > 2.8839 ) = 0.0042

Critical value t(α, DF) = t( 0.05 , 23 ) = 1.7139

Part c-2)

Test Criteria :-
Reject null hypothesis if t > t(α, DF)
t > t(α, DF) = 2.8839 > 1.714
Result :- Reject Null Hypothesis

Decision based on P value
Reject null hypothesis if P value < α level of significance
P - value = 0.0042 < 0.05, hence we reject null hypothesis
Conclusion :- Reject null hypothesis