Tests based on independent samples

(Known variance or large sample size)

Let us consider the case where we have independent random samples from two normally distributed populations. The first population has mean and variance and we obtain a random sample of size . The second population has mean and variance and we obtain a random sample of size .

We know that if the sample means are denoted and , then the random variable

has a standard normal distribution. If the population variances are known, tests for the difference between the population means can be based on this result. The value of the test statistic for is computed as

and the following tests have a significance level

1. To test either null hypothesis

or

against the alternative

the decision rule is

Reject if

2. To test either null hypothesis

or

against the alternative

the decision rule is

Reject if

3. To test the null hypothesis

against the two sided alternative

the decision rule is

Reject if or

Remark: If the sample sizes are large () then a good approximation at significance level can be made if the population variances and are replaced by the sample variances and .

In addition the central limit theorem leads to good approximations even if the populations are not normally distributed.

Example:

According to the Bureau of Labor Statistics, last year university instructors earned an average $440 per month and college instructors earned an average of $420 per month. Assume that these mean earnings have been calculated for samples of 400 and 600 instructors taken from the two populations, respectively. Further assume that the standard deviations of monthly earnings of the two populations are $50 and $63, respectively. Test at 1% significance level if the mean monthly earnings of the two groups of the instructors are different.

Solution:

From the information given above,

; ; ;

; ; ;

where the subscript x refers to university instructors and y -to college instructors. Let

= mean monthly earnings of all university instructors

= mean monthly earnings of all college instructors.

We are to test if the two population means are different. The null and alternative hypotheses are

(the monthly earnings are not different)

(the monthly earnings are different).

The decision rule is

Reject if or

First of all we find the value of . Since , the value of is (approximately) 2.58 and .

The value of the test statistic is computed as follows:

.

and the value of test statistic falls in the rejection region, we reject the null hypothesis . Therefore, we conclude that the mean monthly earnings of the two groups of instructors are different.

Note that we can not say for sure that two means are different. All we can say is that the evidence from the two samples is very strong that the corresponding population means are different.

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