Comparing Averages of Two Paired Samples

To assess the stock performance of various sectors of industry for the year 2002, we look at stock prices for selected companies. Table  7.1 gives stock prices for 6 food service companies and 5 computer industry companies at January of 2002 and 2003.

  

Table 7.1: Stock Prices of Selected Food and Computer Companies

There are two types of mean comparisons allowed by this type of data:

Independent Samples: If we compare food company versus computer company prices (for example 31.16 versus 25.91), this is called a comparison of means between independent samples. The confidence interval procedure for $\mu_1-\mu_2$ is given by either ( 7.7) or ( 7.8).

Paired Samples: If we compare Jan. 2002 prices versus Jan.2003 prices (for example 31.16 versus 28.07), this is called a comparison of means between paired samples . The confidence interval for $\mu_1-\mu_2$ is discussed below.

The best way to tell whether you have independent  or paired samples  is to ask: do I have "two samples" (independent) or "one sample measured twice" ( i.e. paired)? The statistical treatment is different because paired samples involve sampling luck just once. This gives paired-samples better statistical properties than independent samples. For independent samples, luck-of-the-draw occurs two times.

The correct procedure for estimating the difference between means of paired columns is as follows: subtract the two measurements, and use a one-sample confidence interval ( 7.4) or ( 7.5) on the differences.

For the food service companies, the calculations are given in Table  7.2:

  

Table 7.2: Stock Prices of Selected Food Companies

Using ( 7.5), a 95% confidence interval for the difference in means is given by

$$3.09 \pm 2.57 \; (5.74/\sqrt{6})$$

or (-2.93, 9.11).

  

For the independent sample comparison of food versus computer companies (for 2002), the calculation is as follows:

$$S_p= \sqrt{\frac{(6-1)(7.10^2)+(5-1)(6.34^2)}{ 6+5-2} }= 6.77$$

Therefore, $\mbox{SE}_1=6.77/\sqrt{6} = 2.76$, $\mbox{SE}_2=6.77/\sqrt{5}=3.03$, and the difference between means is estimated as

$$31.16 - 25.91 \pm t\; \sqrt{(2.76)^2 + (3.03)^2}$$

where the second term is the standard error. For a 95% confidence interval, the critical value from the t curve with 6+5-2=9 degrees of freedom is 2.26. Therefore a t-confidence interval for $\mu_1-\mu_2$ with confidence level .95 is

$$31.16 - 25.91 \pm 2.26\; \sqrt{(2.76)^2 + (3.03)^2}$$

or (-4.01, 14.51).