Question

Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations.

Sample 1	Sample 2
x¯1=20.92	x¯2=26.80
s21=2.89	s22=3.81
n1=19	n2=15

Test the null hypothesis H0:μ1=μ2against the alternative hypothesis HA:μ1<μ2.

a) Calculate the test statistic for the Welch Approximate t procedure.

Round your response to at least 3 decimal places.

b) The Welch-Satterthwaite approximation to the degrees of freedom is given by df = 27.983055. Using this information, determine the range in which the p-value falls:

		p-value > 0.10
		0.05 < p-value < 0.10
		0.025 < p-value < 0.05
		0.01 < p-value < 0.025
		p-value < 0.01

c) What is the most appropriate conclusion that can be made?

		There is insufficient evidence to reject the null hypothesis at both the 5% and 1% level of significance.
		There is sufficient evidence to reject the null hypothesis at the 1% level of significance, but not at the 5% level of significance.
		There is sufficient evidence to reject the null hypothesis at both the 5% and 1% level of significance.
		There is sufficient evidence to reject the null hypothesis at the 5% level of significance, but not at the 1% level of significance.

Answer 1

The statistical software output for this problem is:

Two sample T summary hypothesis test: H1 Mean of Population 1 H2 Mean of Population 2 H1-2:Difference between two means Ho H-H20 (without pooled variances) Hypothesis test results Difference Sample Diff. Std. Err. P-value DF T-Stat -5.88 0.63726389 27.983055 -9.2269468 <0.0001

Hence,

a) Test statistic = -9.227

b) p-value < 0.01

c) There is sufficient evidence to reject the null hypothesis at both the 5% and 1% level of significance.