If the sample size is small () and the sample distribution is normal or approximately normal, then the Student's t distribution and associated statistics can be used to determine if or test whether the sample mean = population mean.

Comparing sample means of two independent samples with small sample size is similar to comparing a sample mean against a population mean (Chapter 7); the t-statistics or student's t distribution is used to evaluate tests. The only difference is the values for the parameters used in determining the statistics.

The hypothesis testing involving two different means study the distribution of their differences:.

1. Know the basic general statistics used for comparing two population means - small sample size or is unknown.

If we have two populations or sample distributions the following basic statistics can be obtained from each:
 

Population or Sample Identification	Sample size	Sample mean	Population mean	Sample Standard deviation
1	n1
2	n2

Small sample size studies use the student t statistics and large sample sizes studies use the standard normal z-score statistics.

If we let  ( and  and  be a combined standard deviation for both sample distribution or data sets, then
 

For large sample size () the test statistics in a hypothesis test is: , the z-score

For small sample size () the test statistics in a hypothesis test is: , the student's t, df = n-1

For small sample size the standard deviation and test statistics are:
 

Standard Deviation: Pooled sample standard deviation when Also

Test statistics, t, df = n-1 (smallest sample size) Pooled Test statistics, t ,  when  Confidence Interval is

2. Know how to use appropriate statistics to test if two sample means are equal or if their difference = 0 (small sample size).

3 Types of tests in comparing two sample means:

When comparing the sample means,  there are 3 questions to consider:

Question 1: : Is  ? H_a (Two-tailed test)

Question 2: : Is  ? H_a (Right-tailed test)

Question 3: : Is ? H_a (Left-tailed test)

Example 4 is an example of the pool t-test.

Question 1: Is  ? H_a (Two-tailed test)
 

Is Ha Is the two sample means or the same or not the same?  If the two means are the same then their difference is approximately = 0. Here we use a two-tailed test by computing the confidence interval for the test at the level of significance . If the test statistics for z falls outside this interval we decide that the means differs, we chose the alternate hypothesis, Ha Otherwise we have no reason to think that they differ, H0

By Examples:

Problem 1. Two types of cars are compared for acceleration rate. The test runs are recorded for each car and the results for the mean elapsed time recorded below:
 

	Sample mean	Sample standard deviation	Sample size
Car A (x1)	8.5	1.8	20
Car B (x2)	7.2	2.1	30

Construct a 98% CI for the difference in the mean elapsed time for the two types of cars. Using this CI, determine if there is a difference in the mean elapsed times?

Given difference , at least one of the same < 30 (small so must use the student's t distribution or t-statistics), 

Step 1 - Hypothesis: The claim that  , the null hypothesis.

The alternate hypothesis is that 

H₀ :

H_a :

Step 2. Select level of significance: This is given as  (2% = 100 - 98) 

So for two-tailed test: 

Step 3. Test statistics and observed value.

Step 4. Determine the critical region (favors H_a)

For alpha = 0.02 at both ends of intervals: 0.01 and 0.99, t_a = 2.5395 and -t_a = -2.5395 (from reference table)
 

The critical region is  and  with df = 19; smallest of (20-1) and (30-1)

A 98% Confidence Interval for the difference is  or (diff: -0.1117 to 2.7117) shown in graph above

Step 5. Make decision.

No not reject the null hypothesis if  or 

The observed t=2.3386, and since 2.3386 < 2.5395 and is not in the critical region, we have no reason to reject H₀ in favor of H_a.

Note also that a difference of 1.3 is between the confidence intervals of -0.1117 and 2.7117 the blue region for the null hypothesis acceptance.

Therefore the difference between both means are significantly different from 0.

Question 2: : Is  ? H_a (Right-tailed test)
 

Is Ha Is the sample mean of sample 1 greater than the sample mean of sample 2?  This test evaluate if their difference will be significantly greater than zero or not equal to 0. Here we use a one-tailed test by computing the confidence interval for the test at the level of significance . If the test statistics for z falls outside this interval we decide that the means differs, we chose the alternate hypothesis, Ha Otherwise we have no reason to think that they differ, H0

By Examples:

Problem 2. The personnel officer of a large corporation claimed that college graduates applying for jobs with their firm in the current year tended to have higher grade point averages than those applying in the previous year. Samples from the group of applicants gave the following results:
 

	Sample mean	Sample standard deviation	Sample size
Current year (x1)	3.21	0.51	20
Previous year (x2)	2.95	0.42	30

Is there sufficient evidence to justify the claim at a 5% level of significance?

Given difference , at least one of the sample size < 30 (small so must use the student's t distribution or t-statistics),

Step 1 - Hypothesis: The claim that  , the null hypothesis.

The alternate hypothesis is that 

H₀ :

H_a :

Step 2. Select level of significance: This is given as  (5%) 

Step 3. Test statistics and observed value.

Step 4. Determine the critical region (favors H_a)

For alpha = 0.05 at the upper end of intervals: 0.95, t_a = 1.7291  (from reference table)
 

The critical region is  with df = 19; smallest of (20-1) and (30-1) reference table

Step 5. Make decision.

No not reject the null hypothesis if 

Since 1.892 is > 1.7291 and in the critical region (red) we reject the null hypothesis that grades are the same both years.

The observed t = 1.89 and since 1.89 > 1.73 and is in the critical region, we reject H₀ in favor of H_a.

Therefore we conclude that college graduates from current year have higher grades than previous year.

Question 3: : Is ? H_a (Left-tailed test)
 

Is Ha Is the sample mean of sample 1 less than the sample mean of sample 2?  If so then their difference will be significantly greater than zero or not equal to 0. Here we use a one-tailed test by computing the confidence interval for the test at the level of significance . If the test statistics for z falls outside this interval we decide that the means differs, we chose the alternate hypothesis, Ha Otherwise we have no reason to think that they differ, H0

By Examples:

Problem 3. A biologist suspected that females age 20 - 24 have a lower mean systolic blood pressure than males in the same age group. Independent random sample produced the following results for systolic pressure.
 

	Sample mean	Sample standard deviation	Sample size
Female (x1)	114	12.5	14
Male (x2)	122	13.9	21

Is there sufficient evidence to justify the claim at a 1% level of significance?

Given difference , at least one of the sample size < 30 (small so must use the student's t distribution or t-statistics),

Step 1 - Hypothesis: The claim that  , the null hypothesis.

The alternate hypothesis is that 

H₀ :

H_a :

Step 2. Select level of significance: This is given as  (5%) 

Step 3. Test statistics and observed value.

Step 4. Determine the critical region (favors H_a)

For alpha = 0.05 at the upper end of intervals: 0.95, -t_a =-1.7709  (from reference table)
 

The critical region is  with df = 13; smallest of (14-1) and (21-1)

Step 5. Make decision.

No not reject the null hypothesis if 

Since -1.7729 is < -1.7709 and in the critical region (red) we reject the null hypothesis that female and male systolic pressure are the same..

The observed t = -1.7729 and since -1.7729 < 1.7709 and is in the critical region, we reject H₀ in favor of H_a.

Therefore we conclude that female systolic pressure are lower than male's same age (20-24)

Pooled t-test:  (Question 4)

If samples being compared are from the same population where , then the simplified pooled statistics can be used to evaluate the test of two samples means.

Some statistician like myself would rather used the above t-tests when regardless of whether .
 

Pooled Test statistics, t ,  when  Confidence Interval is

Pooled Standard deviation

Pooled t procedure Is  ? H_a (one-tailed test)
 

Is Ha Is the sample mean of sample 1 less than the sample mean of sample 2?  If so then their difference will be significantly greater than zero or not equal to 0. Here we use a one-tailed test by computing the confidence interval for the test at the level of significance . If the test statistics for z falls outside this interval we decide that the means differs, we chose the alternate hypothesis, Ha Otherwise we have no reason to think that they differ, H0

By Examples: Pooled t-procedure

Problem 4. Assume that the populations of both Test A and Test B below are approximately normal with same or equal standard deviations. Is the mean of Test A less than the mean of Test B (compare at a 5% level of significance).
 

	Sample mean	Sample standard deviation	Sample size
Test A (x1)	120	10	6
Test B (x2)	125	9	10

Given difference , at least one of the same < 30 (small so must use the student's t distribution or t-statistics), and since assuming that , we will use the pooled t-test.

Step 1 - Hypothesis: The claim that , the null hypothesis.

The alternate hypothesis is that 

H₀ :

H_a :

Step 2. Select level of significance: This is given as  (5%) 

Step 3. Test statistics and observed value. (since pooled we used the improved pooled statistics:

and 

when 

Step 4. Determine the critical region (favors H_a)

For alpha = 0.05 at the upper end of intervals: 0.95, -t_a =-1.7613  (from reference table)
 

The critical region is  with df = 14; n1+n2-2 = 6 + 10 - 2=14

Step 5. Make decision.

No not reject the null hypothesis if 

Since -1.0047 is > -1.7613 and in not in the critical region (red) we have no reason to reject the null hypothesis that both means are the same.

The observed t = -1.0047 and since -1.0047 > -1.7613 and is not in the critical region, we accept H₀.

Therefore we conclude that means are the same for both Tests (or no significant difference between them).