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-criticaln =

(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 = 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 =

(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 = 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 =

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

• Reject the null hypothesis

• Do not reject the null hypothesis

Answer 1

(a-1) Comparison of GPA for randomly chosen college juniors and seniors

The null and alternative hypothesis is given by

Ho : There is no significant difference in the comparison of GPA for randomly chosen college juniors and seniors.

H1 : There is significant difference in the comparison of GPA for randomly chosen college juniors and seniors.

The values provided in the above question are as below

Degrees of freedom = n1 + n2 - 2 = 15 + 15 - 2 = 28

Degrees of freedom = d.f. = 28

The formula of t-calculated is as

  

t-calculated = -4.6189

We find p-value using Excel function

=TDIST(x,deg_freedom,tails)

here, x = t-calculated = -4.6189 but consider positive sign = 4.6189, deg_freedom = 28, tails = 1

=TDIST(4.6189,28,1) then press Enter

p-value = 0.0000392878 (We write value using 10 decimal places)

t-critical using t table at degrees of freedom 28 and corresponding level of significance = = 0.025

We take negative sign for t-critical because the test is left tailed test

t-critical = -2.048

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

Here, p-value = 0.0000392878 level of significance = = 0.025

So, We reject the null hypothesis.

That is, there is significant difference in the comparison of GPA for randomly chosen college juniors and seniors.

Summary :-

(a-1) Comparison of GPA for randomly chosen college juniors and seniors

d.f. = 28

t-calculated = -4.6187

p-value = 0.0000392878

t-critical = -2.048

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

Reject the null hypothesis.

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

The null and alternative hypothesis is given by

Ho : There is no significant difference in the comparison of average commute miles for randomly chosen students at two community colleges.

H1 : There is significant difference in the comparison of average commute miles for randomly chosen students at two community colleges.

The values provided in the above question are as below

(Two-tailed test)

Degrees of freedom = n1 + n2 - 2 = 22 + 19 - 2 = 39

Degrees of freedom = d.f. = 39

The formula of t-calculated is as

  

t-calculated = -4.1504

We find p-value using Excel function

=TDIST(x,deg_freedom,tails)

here, x = t-calculated = -4.1504 but consider positive sign = 4.1504, deg_freedom = 39, tails = 2

=TDIST(4.1504,39,2) then press Enter

p-value = 0.000174211 (We write value using 9 decimal places)

t-critical using Excel function at degrees of freedom 39 and corresponding level of significance = = 0.05 (two tailed test)

=TINV(probability,deg_freedom)

=TINV(0.05,39)

We take sign for t-critical because the test is two tailed test

t-critical = 2.022691  

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

Here, p-value = 0.000174211 level of significance = = 0.05

So, We reject the null hypothesis.

That is, there is significant difference in the comparison of average commute miles for randomly chosen students at two community colleges.

Summary :-

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

d.f. = 39

t-calculated = -4.1504

p-value = 0.000174211

t-critical = 2.022691

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

Reject the null hypothesis.

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

The null and alternative hypothesis is given by

Ho : There is no significant difference in the comparison of credits at time of graduation for randomly chosen accounting and economics student.

H1 : There is significant difference in the comparison of credits at time of graduation for randomly chosen accounting and economics student.

Using the values provided in the above question are as below

x¯1 = 150, s1 = 2.8, n1 = 12, x¯2 = 143, s2 = 2.7, n2 = 17, α = .05, right-tailed test

Degrees of freedom = n1 + n2 - 2 = 12 + 17 - 2 = 27

Degrees of freedom = d.f. = 27

The formula of t-calculated is as

Using all values we get

t-calculated = 6.729

Using Excel function (=TDIST(x,deg_freedom,tails))

=TDIST(6.729,27,1) then press Enter

p-value = 0.0000001588 (We write value using 10 decimal places)

t-critical using t table at degrees of freedom 27 and corresponding level of significance = = 0.05(one-tailed test)

t-critical = 1.703

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

Here, p-value = 0.0000001588 level of significance = = 0.05

So, We reject the null hypothesis.

That is, there is significant difference in the comparison of credits at time of graduation for randomly chosen accounting and economics student.

Summary :-

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

d.f. = 27

t-calculated = 6.729

p-value = 0.0000001588

t-critical = 1.703

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

Reject the null hypothesis