TEST OF HYPOTHESIS ABOUT MU1

TEST OF HYPOTHESIS ABOUT m₁ – m₂

INDEPENDENT SAMPLES

It is conjectured that classes that use a statistical computer package, such as Minitab, do better in introductory statistics courses than those who don’t use such technology. A random sample of 24 students uses a statistical computer package while taking statistics. Another random sample of 28 students taking the same course uses only hand-held calculators. The final average in the course is recorded for each of these students. These data are below.

Is there sufficient evidence to conclude that students who do not use the computer have lower averages? Use a = .05.

SOLUTION

The parameter of interest is m₁ – m₂, the difference in the average final average of students using the computer versus the average final average of students not using the computer.

H₀: m₁ – m₂ = 0 H_a: m₁ – m₂ > 0

Decision Rule: Accept H_a if the calculated p-value < .05.

Calculations from StatCrunch: t = 1.80, p-value = 0.0392 < .05 ---> Accept H_a

Interpretation: At the .05 level of significance I conclude that the true mean final average of students using the computer in this statistics course is higher than those not using the computer.

COMMENTS ABOUT THE SOLUTION

The usual structure for presenting the solution to a test of hypothesis was followed. All test of hypotheses solutions should follow this structure.

Since two different random samples were selected, the samples are independent.

This procedure is valid when both populations are normally distributed, both samples are randomly selected, and the two samples are independent from each other.

The procedure in this problem is also valid when large (30 or more) samples are randomly selected from any population, regardless of its original distribution.

The complete StatCrunch analyses are below.

Hypothesis test results:
μ₁ : mean of Computer
μ₂ : mean of No Computer
μ₁ - μ₂ : mean difference
H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ > 0
(without pooled variances)

Difference

Sample Mean

Std. Err.

DF

T-Stat

P-value

μ₁ - μ₂

4.821429

2.6779594

45.597992

1.8004113

0.0392

A graph of the appropriate t-distribution and the resulting p-value is furnished below.

Difference	Sample Mean	Std. Err.	DF	T-Stat	P-value
μ₁ - μ₂	4.821429	2.6779594	45.597992	1.8004113	0.0392