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 = 4.05, s₁ = .20, n₁ = 15, x⎯⎯2x¯2 = 4.35, 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 = 19, s₁ = 5, n₁ = 22, x⎯⎯2x¯2 = 25, 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.
  

• Reject the null hypothesis

• Do not reject the null hypothesis

(c-1) Comparison of credits at time of graduation for randomly chosen accounting and economics students:

x⎯⎯1x¯1 = 144, s₁ = 2.8, n₁ = 12, x⎯⎯2x¯2 = 143, 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)

Test Statistic :-


t = -3.2225

Test Criteria :-
Reject null hypothesis if t < -t(α, DF)


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 = -3.1128

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, DF)


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 = 0.9613

Test Criteria :-
Reject null hypothesis if t > t(α, DF)


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