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 = 4.75, s₁ = .20, n₁ = 15, x⎯⎯2x¯2 = 5.18, 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

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

x⎯⎯1x¯1 = 25, s₁ = 5, n₁ = 22, x⎯⎯2x¯2 = 33, 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

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

x⎯⎯1x¯1 = 150, 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

Answer 1

1) df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

       = ((0.2)^2/15 + (0.3)^2/15)^2/(((0.2)^2/15)^2/14 + ((0.3)^2/15)^2/14)

        = 24

The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                             = (4.75 - 5.18)/sqrt((0.2)^2/15 + (0.3)^2/15)

                             = -4.6189

P-value = P(T < -4.6189)

             = 0.0001

At alpha = 0.025, the critical value is t^* = -2.064

b) df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

       = (5^2/22 + 7^2/19)^2/((5^2/22)^2/21 + (7^2/19)^2/18)

       = 32

The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                             = (25 - 33)/sqrt(5^2/22 + 7^2/19)

                             = -4.1504

P-value = 2 * P(T < -4.1504)

             =2 * 0.0001 = 0.0002

At alpha = 0.05, the critical values are t^* = +/- 2.037

c) df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

       = ((2.8)^2/12 + (2.7)^2/17)^2/(((2.8)^2/12)^2/11 + ((2.7)^2/17)^2/16)

      = 23

The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                             = (150 - 143)/sqrt((2.8)^2/12 + (2.7)^2/17)

                             = 6.7290

P-value = P(T > 6.7290)

             = 1 - P(T < 6.7290)

              = 1 - 1 = 0

At alpha = 0.05, the critical value is t^* = 1.714