INFERENCE FOR THE MEANS OF TWO POPULATIONS

A. Remember the special one-sample example

1. Example (matched pairs t test)

   In an experiment done by Mark Rosenzweig et al., a pair of rats were chosen from 11 litters. In each pair, one rat was given playthings; the other rat was isolated with no toys. After a month, the animals were dissected and their brain cortexes were weighed. For most of the pairs, the treatment rat had a heavier cortex:

   32 33 16 6 21 17 64 7 89 -2 -9

   If playthings made no difference, we would expect the brain cortexes to have the same weight, so the expected difference would be zero. Since we don't expect every difference to be zero, we suspect the true differences would follow a normal curve with a mean of zero and with an unknown SD. Since this is a randomized experiment, these data can be treated as a simple random sample.

   Does it look like playthings make a difference in cortex weight?

   H0: the differences come from a normal curve with mean zero

   Ha: the differences come from a normal curve with a mean greater than zero

   The mean of the list is 24.91; the SD of the sample is 29.09. The estimated SD of the population is thus 29.09. The estimated standard error is 29.09/sqrt(11) = 8.77 with 10 df.

   Now, t = (24.91-0.00)/8.77 = 2.84 with 10 df. According to Table C, this has a p-value between .01 and .005.

   Therefore, reject the null hypothesis: the true mean difference looks like it is greater than zero (i.e., playthings make a difference).

B. Background for the Two Sample Case

In the matched-pairs case, even though their are two groups, they are treated as one because the groups are not independent. In other words, one group determines what the other will look like. Matched pair t-tests are frequently "before and after" comparisons or experiments when subjects are matched totally (e.g. identical twins)

Now you are in the situation where we have two groups and you are interested in comparing their means. You can think of these two groups as simple random samples from two distinct and independent populations and you wish to compare some characteristic between them.

There are (a) two-sample Z statistic and (b) two-sample T-statistic

C. Two-sample Z

A simple random sample of 100 UCLA students had an average IQ of 104 with an SD of 16. An independent simple random sample of 100 University of Southern California students had an average IQ of 98, with an SD of 14. Could this 6 IQ point difference be explained by the luck of the draw, or does this show that the average IQ at UCLA is higher than the average IQ at USC?

Recall:

• The null hypothesis is that the observed results are due to chance.
• The p-value measures the chance of getting the observed (or more extreme) results, if the null hypothesis were true.
• If p< alpha (some level of significance) the null hypothesis is rejected.

D. Method 1: a Two Sample Z

1. The following method can be used when drawing independent simple random samples from two populations and either of the following two conditions is true:

   a. both populations are known to be normally distributed and the population SD's are known; or,

   b. the sample sizes are "large."

2. The null hypothesis is that the two populations have the same means, so the expected difference is zero.

3. The observed difference in means is simply the difference in sample means, xbar1 - xbar2. For the current example, the observed difference is 6 IQ points.

4. The standard deviation for the difference is

   sqrt( sigma1^2/n1 + sigma2^2/n2 ),

   where sigma_1 and sigma_2 are the standard deviations for the first and second populations, and n_1 and n_2 are the sizes of the two samples.

   In the current example, the standard deviation for the difference is 2.13.

5. By assumption, the difference xbar1 - xbar2 will have a normal distribution. Thus, we can use the normal curve on the difference

   z = (xbar1 - xbar2) - (mu1 - mu2)/sqrt( sigma1^2/n1 + sigma2^2/n2 ),

   Thus, if the null hypothesis were true, we would expect to observe a difference of zero, give or take 2.13 or so. (Draw a picture!) Then z = (6-0)/2.13 = 2.82, and p is 0.2 of 1%.

   We thus would reject the null hypothesis: it seems the average IQ at UCLA is higher than the average IQ at USC.

E. Method 2: A two sample t test

1. The following method should be used when both populations are known to be normally distributed, the population SD's are unknown, and independent simple random samples are drawn from the two populations. (This method is only used when samples are small; for example, if the two samples were of size 10 and 8 instead of 100 and 100.)

2. As before, the null hypothesis is that the population means are the same; i.e., that the expected difference is zero.

3. The standard error is estimated as

std error = sqrt( SD1^2/n1 + SD2^2/n2 ),

where SD1 is the SD of the first sample, and SD2 is the SD of the second sample. If the sample sizes were 10 and 8, the standard error for the difference would be 7.08.

4. The appropriate distribution is t with degrees of freedom equal to the smaller of the two sample sizes minus one. In the current example, the degrees of freedom would be 7.

5. If the null hypothesis were true, we would expect a difference of zero, give or take about 7.08. (Draw a picture!)

Then t = (6 - 0)/7.08 = 0.847 with 8-1 = 7 degrees if freedom.

The p-value is thus between 20% and 25% (the actual p-value is 21% but you don't need to know this): inconclusive. Here, we could NOT reject the null hypothesis.