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⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 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

(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⎯⎯1x1 = 25, s1 = 5, n1 = 22, x⎯⎯2x2 = 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⎯⎯1x1 = 150, s1 = 2.8, n1 = 12, x⎯⎯2x2 = 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

Solution:-

a)

a-1)

Comparison of GPA for randomly chosen college juniors and seniors.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁> u₂
Alternative hypothesis: u₁ < u₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.025. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.0931
DF = 28

t = [ (x₁ - x₂) - d ] / SE

t = - 4.62

t_critical = -2.049

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of -4.62.

Therefore, the P-value in this analysis is less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.025), we have to reject the null hypothesis.

a-2) Reject the null hypothesis.