Comparison of Two Means

In many cases, a researcher is interesting in gathering information about two populations in order to compare them. As in statistical inference for one population parameter, confidence intervals and tests of significance are useful statistical tools for the difference between two population parameters.

Confidence Interval for the Difference Between Two Means

A confidence interval for the difference between two means specifies a range of values within which the difference between the means of the two populations may lie. These intervals may be calculated by, for example, a producer who wishes to estimate the difference in mean daily output from two machines; a medical researcher who wishes to estimate the difference in mean response by patients who are receiving two different drugs; etc. The confidence interval for the difference between two means contains all the values of ( - ) (the difference between the two population means) which would not be rejected in the two-sided hypothesis test of
H0: = against Ha: , i.e.
H0: - = 0 against Ha: - 0.

If the confidence interval includes 0 we can say that there is no significant difference between the means of the two populations, at a given level of confidence.

(Definition taken from Valerie J. Easton and John H. McColl's Statistics Glossary v1.1)


Tests of Significance for Two Unknown Means and Known Standard Deviations

Given samples from two normal populations of size n1 and n2 with unknown means and and known standard deviations and , the test statistic comparing the means is known as the two-sample z statistic

which has the standard normal distribution (N(0,1)).

The null hypothesis always assumes that the means are equal, while the alternative hypothesis may be one-sided or two-sided.


Tests of Significance for Two Unknown Means and Unknown Standard Deviations

In general, the population standard deviations are not known, and are estimated by the calculated values s1 and s2. In this case, the test statistic is defined by the two-sample t statistic
.
Although the two-sample statistic does not exactly follow the t distribution (since two standard deviations are estimated in the statistic), conservative P-values may be obtained using the t(k) distribution where k represents the smaller of n1-1 and n2-1. Another option is to estimate the degrees of freedom via a calculation from the data, which is the general method used by statistical software such as MINITAB.

The confidence interval for the difference in means - is given by

where t* is the upper (1-C)/2 critical value for the t distribution with k degrees of freedom (with k equal to either the smaller of n1-1 and n1-2 or the calculated degrees of freedom).

Example

The dataset "Normal Body Temperature, Gender, and Heart Rate" contains 130 observations of body temperature, along with the gender of each individual and his or her heart rate. In the dataset, the first column gives body temperature and the second column gives the value "1" (male) or "2" (female) to describe the gender of each subject. Using the MINITAB "DESCRIBE" command with the "BY" subcommand to separate the two genders provides the following information:
Descriptive Statistics

Variable  C2              N     Mean   Median  Tr Mean    StDev  SE Mean
C1        1              65   98.105   98.100   98.114    0.699    0.087
          2              65   98.394   98.400   98.390    0.743    0.092

Variable  C2            Min      Max       Q1       Q3
C1        1          96.300   99.500   97.600   98.600
          2          96.400  100.800   98.000   98.800
Is there a significant difference between the mean body temperatures for men and women? To test H0: - = 0 against Ha: - 0, compute the test statistic (98.105 - 98.394)/(sqrt(0.699²/65 + 0.743²/65)) = -0.289/0.127 = -2.276. Using the t(64) distribution, estimated in Table E in Moore and McCabe by the t(60) distribution, we see that 2P(t>2.276) is between 0.04 and 0.02, indicating a significant difference between the means at the 0.05 level (although not at the 0.01 level).

To compute a 95% confidence interval, we first note that the 0.025 critical value t* for the t(60) distribution is 2.000, giving the interval ((98.105 - 98.394) + 2.000*0.127) = (-0.289 - 0.254, -0.289 + 0.254) = (-0.543, -0.045). The value 0 is not included in the interval, again indicating a significant difference at the 0.05 level.

Performing this test in MINITAB using the "TWOT" command gives the results

Two Sample T-Test and Confidence Interval

Two sample T for C1
C2           N      Mean     StDev   SE Mean
1           65    98.105     0.699     0.087
2           65    98.394     0.743     0.092

95% CI for mu (1) - mu (2): ( -0.540,  -0.039)
T-Test mu (1) = mu (2) (vs not =): T= -2.29  P=0.024  DF=  127  
Although the MINITAB calculated degrees of freedom (127) are much higher than the conservative estimate of 64, we see that the results are much the same.

Data source: Data presented in Mackowiak, P.A., Wasserman, S.S., and Levine, M.M. (1992), "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich," Journal of the American Medical Association, 268, 1578-1580. Dataset available through the JSE Dataset Archive.


Pooled t Procedures

If it reasonable to assume that two populations have the same standard deviation, than an alternative procedure known as the pooled t procedure may be used instead of the general two-sample t procedure. Since only one standard deviation is to be estimated in this case, the resulting test statistic will exactly follow a t distribution with n1 + n2 - 2 degrees of freedom. The pooled estimator of the variance
is used in the pooled two-sample t statistic
which has a t(n1 + n2 -2) distribution.

Example

In the body temperature example above, the sample standard deviations for the male and female subjects are reasonable close. Using the MINITAB subcommand "POOLED" with the two-sample t test gives the following results:
Two Sample T-Test and Confidence Interval

Two sample T for C1
C2           N      Mean     StDev   SE Mean
1           65    98.105     0.699     0.087
2           65    98.394     0.743     0.092

95% CI for mu (1) - mu (2): ( -0.540,  -0.039)
T-Test mu (1) = mu (2) (vs not =): T= -2.29  P=0.024  DF=  128
Both use Pooled StDev = 0.721         
The test results were nearly identical in this case.