Appropriate test statistic and hypothesis test concerning the equality of the population means of two at least approximately normally distributed populations based on independent random samples with equal or unequal assumed variances

CFA level I / Quantitative Methods: Application / Hypothesis Testing / Appropriate test statistic and hypothesis test concerning the equality of the population means of two at least approximately normally distributed populations based on independent random samples with equal or unequal assumed variances

Sometimes the researchers need to test whether the mean of two different normally distributed populations is different due to chance or to different underlying values of mean. We have t-tests to check this. There are two t-tests available depending on whether the unknown population variances are assumed equal or unequal.

The hypotheses tests in testing the equality of means would be following:

H0: µ1 - µ2 = 0 versus Ha: µ1 - µ2 ≠ 0 (two-tailed test)
H0: µ1 - µ2 ≤ 0 versus Ha: µ1 - µ2 > 0 (right-tailed test)
H0: µ1 - µ2 ≥ 0 versus Ha: µ1 - µ2 < 0 (left-tailed test)

Population variances are unknown and assumed equal:

t = [(X ̅1 - X ̅2) - (µ1 - µ2)]/[s2p/n1 + s2p/n2]0.5

where s2p = [(n1 - 1)s21 + (n2 - 1)s22]/(n1 + n2  - 2)

where s2p is the pooled estimator of the common variance

µ1 and µ2 are population means of first and second populations

n1 and n2 are sample sizes of first and second populations

The number of degrees of freedom equals n1 + n2  - 2.

Example 3: Testing equality of population means with unknown

You want to test whether the monthly returns in the first half of the year is equal to the second half of the year. You take a sample of 32 months from the first half with a mean of 1.10 percent and standard deviation of 1.90 percent. The sample mean and standard deviation of the observations from the second half of the year are 0.90 percent and 1.80 percent respectively for a total of 50 observations. Assuming the population variances of two populations to be same, check whether the population means are equal or not at 1 percent level of significance.

Solution:

The hypotheses are given below for the t-test:

Null hypothesis, H0: µ1 - µ2 = 0
Alternative hypothesis, Ha: µ1 - µ2 ≠ 0

It is a two-tailed test. So, the null hypothesis could be rejected in both the tails. The t-critical value will be ±t0.005 for degrees of freedom of 80 (=32+50-2). The t-critical values are -2.639 and 2.639.

s2p = [(n1 - 1)s21 + (n2 - 1)s22]/(n1 + n2  - 2) = [31*0.0192 + 49*0.0182]/(32+50-2) = 0.000338.

t-statistic = [(X ̅1 - X ̅2) - (µ1 - µ2)]/[s2p/n1 + s2p/n2]0.5 = (0.011 - 0.009)/[(0.000338/32) + (0.000338/50)]0.5 = 0.4803.

Since, the test statistic value of 0.4803 lies between the critical values of -2.639 and +2.639, we fail to reject the null hypothesis.

The result of the test is that the population means are statistically equal at 1 percent level of significance.

Please note that even without calculating the degrees of freedom, we could have deduced the same result because the smallest critical values for 1 percent are ±2.576 (when the number of degrees of freedom is equal to infinite) and 0.4803 lies between ±2.576.

Population variances are unknown and assumed unequal:

t = [(X ̅1 - X ̅2) - (µ1 - µ2)]/[s21/n1 + s22/n2]0.5

Degrees of freedom, df = [s21/n1 + s22/n2]2/[{(s21/n1)2/n1} + {(s22/n2)2/n2}]

One should calculate the test statistic first before computing the degrees of freedom. We might fail to reject the null hypothesis for a small test statistic value without even looking at the degrees of freedom.

Example 4: Testing equality of population means with unknown and unequal variances

You want to test whether the recovery rate on the defaulted bonds is same for the chemical sector and automobile sector. For the hypotheses testing purpose, you take a sample of 15 and 25 defaulted bonds for chemical and automobile sector respectively and check their recovery rates. The mean recovery rates for samples for chemical and automobile sectors are 0.55 and 0.42 respectively. The standard deviation of the recovery rates for chemical and automobile sectors are 0.17 and 0.19 respectively. The population variances are assumed to be unequal for the recovery rates of the given sectors. Formulate the null and alternative hypotheses and test the hypothesis of the equality of population means at 5 percent level of significance.

Solution:

The hypotheses are given below for the t-test:

Null hypothesis, H0: µ1 - µ2 = 0
Alternative hypothesis, Ha: µ1 - µ2 ≠ 0

where µ1 and µ2 are population means for recovery rates for chemical and automobile sector respectively.

The test is a two-tailed test.

Degrees of freedom, df = [s21/n1 + s22/n2]2/[{(s21/n1)2/n1} + {(s22/n2)2/n2}] = [0.172/15 + 0.192/25]2/[{(0.172/15)2/15} + {(0.192/25)2/25}] = 34.34.

The critical test statistic value for 34 degrees of freedom at 5 percent level of significance for a two-tailed test are -2.032 and 2.032.

t-statistic = [(X ̅1 - X ̅2) - (µ1 - µ2)]/[s21/n1 + s22/n2]0.5  = (0.55 - 0.42)/[ 0.172/15 + 0.192/25]0.5 = 2.24.

The calculated t-statistic value does not lie between the critical value of -2.032 and 2.032. The value lies in the alternative hypothesis region. So, we reject the null hypothesis.

The recovery rates of chemical and automobile sectors are statistically not equal at 5 percent level of significance.

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