Question

Researchers investigating the ages of houses selected a random sample of houses in Whiting, Indiana and Franklin, Pennsylvania and obtained the following summary statistics.

Whiting	Franklin
x1 = 61.8 yrs	x2 = 57.6 yrs
s1 = 5.9 yrs	s2 = 5.6 yrs
n1 = 20	n2 = 22

Can it be concluded that houses in Whiting are older on average than in Franklin?

Estimate the P-value for a test of this claim.

0.01 < P-value < 0.025

0.02 < P-value < 0.05

P-value < 0.005

0.01 < P-value < 0.02

P-value < 0.01

Answer 1

using excel>addin>phstat>two sample test

we have

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.05
Population 1 Sample
Sample Size	20
Sample Mean	61.8
Sample Standard Deviation	5.9
Population 2 Sample
Sample Size	22
Sample Mean	57.6
Sample Standard Deviation	5.6
Intermediate Calculations
Population 1 Sample Degrees of Freedom	19
Population 2 Sample Degrees of Freedom	21
Total Degrees of Freedom	40
Pooled Variance	32.99875
Difference in Sample Means	4.2
t Test Statistic	2.366477
Upper-Tail Test
Upper Critical Value	1.683851
p-Value	0.011444
Reject the null hypothesis

p value is 0.011

0.01 < P-value < 0.025