Question

Suppose you are a loan office for a bank, and you want to compare interest rates on first mortgages at your branches last month. YOu collect the following data: number 21, 41 mean interest rate 6.60% , 5.90% standard deviation 0.35%, 0.28% . Set up the hypothesis with the proper H0 and H1. Find the F critical value for a=0.05

Answer 1

Before going to test the above hypothesis, first we check the assumption of equal variances are valid or not.

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05

Statistics

Sample N StDev Variance
1 21 0.350 0.122
2 41 0.280 0.078

Ratio of standard deviations = 1.250
Ratio of variances = 1.562

95% Confidence Intervals

CI for
Distribution CI for StDev Variance
of Data Ratio Ratio
Normal (0.869, 1.890) (0.756, 3.574)

Tests

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 20 40 1.56 0.226

From above test we see that p-value>0.05 (or we can also check this by using critical value i.e. value of F statistic=1.56>Critical value=F_0.05,20,40=1.84) hence we assume the population variances are same.

Now we apply 2 sample t test with equal variances.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 21 6.600 0.350 0.076
2 41 5.900 0.280 0.044

Difference = mu (1) - mu (2)
Estimate for difference: 0.7000
95% CI for difference: (0.5362, 0.8638)
T-Test of difference = 0 (vs not =): T-Value = 8.55 P-Value = 0.000 DF = 60
Both use Pooled StDev = 0.3051

Since p-value=0.000<0.05 so  interest rates on first mortgages at the branches last month are significantly different.