Answer:
Calculated t=2.42
Given Df = 65
Critical value of t with 65 DF at 0.05 level = 1.997
Rejection Region: Reject Ho if t < -1.997 or t > 1.997
Calculated t = 2.42 falls in the rejection region
The null hypothesis is rejected.
There is sufficient evidence to conclude that there is significant difference in the mean distance travel per day between Houston and Seattle.
Separate-Variances t Test for the Difference Between Two Means |
|
(assumes unequal population variances) |
|
Data |
|
Hypothesized Difference |
0 |
Level of Significance |
0.05 |
Population 1 Sample |
|
Sample Size |
60 |
Sample Mean |
25.5 |
Sample Standard Deviation |
7.9000 |
Population 2 Sample |
|
Sample Size |
35 |
Sample Mean |
21.1 |
Sample Standard Deviation |
8.9000 |
Intermediate Calculations |
|
Numerator of Degrees of Freedom |
10.9119 |
Denominator of Degrees of Freedom |
0.1690 |
Total Degrees of Freedom |
64.5749 |
Degrees of Freedom |
64 |
Standard Error |
1.8175 |
Difference in Sample Means |
4.4000 |
Separate-Variance t Test Statistic |
2.4209 |
Two-Tail Test |
|
Lower Critical Value |
-1.9977 |
Upper Critical Value |
1.9977 |
p-Value |
0.0183 |
Reject the null hypothesis |
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