Let's start with the good news, namely that we've already done the dirty theoretical work in developing a hypothesis test for the difference in two population means \(\mu_1-\mu_2\) when we developed a \((1-\alpha)100\%\) confidence interval for the difference in two population means. Recall that if you have two independent samples from two normal distributions with equal variances \(\sigma^2_X=\sigma^2_Y=\sigma^2\), then:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}}\)

follows a \(t_{n+m-2}\) distribution where \(S^2_p\), the pooled sample variance:

\(S_p^2=\dfrac{(n-1)S^2_X+(m-1)S^2_Y}{n+m-2}\)

is an unbiased estimator of the common variance \(\sigma^2\). Therefore, if we're interested in testing the null hypothesis:

\(H_0:\mu_X-\mu_Y=0\) (or equivalently \(H_0:\mu_X=\mu_Y\))

against any of the alternative hypotheses:

\(H_A:\mu_X-\mu_Y \neq 0,\quad H_A:\mu_X-\mu_Y < 0,\text{ or }H_A:\mu_X-\mu_Y > 0\)

we can use the test statistic:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}}\)

and follow the standard hypothesis testing procedures. Let's take a look at an example.

Let's again start with the good news that we've already done the dirty theoretical work here. Recall that if you have two independent samples from two normal distributions with unequal variances \(\sigma^2_X \neq \sigma^2_Y\), then:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{\sqrt{\dfrac{S^2_X}{n}+\dfrac{S^2_Y}{m}}}\)

follows, at least approximately, a \(t_r\) distribution where \(r\), the adjusted degrees of freedom is determined by the equation:

\(r=\dfrac{\left(\dfrac{s^2_X}{n}+\dfrac{s^2_Y}{m}\right)^2}{\dfrac{(s^2_X/n)^2}{n-1}+\dfrac{(s^2_Y/m)^2}{m-1}}\)

If r doesn't equal an integer, as it usually doesn't, then we take the integer portion of \(r\). That is, we use \(\lfloor r\rfloor\) if necessary.

With that now being recalled, if we're interested in testing the null hypothesis:

\(H_0:\mu_X-\mu_Y=0\) (or equivalently \(H_0:\mu_X=\mu_Y\))

against any of the alternative hypotheses:

\(H_A:\mu_X-\mu_Y \neq 0,\quad H_A:\mu_X-\mu_Y < 0,\text{ or }H_A:\mu_X-\mu_Y > 0\)

we can use the test statistic:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{\sqrt{\dfrac{S^2_X}{n}+\dfrac{S^2_Y}{m}}}\)

and follow the standard hypothesis testing procedures. Let's return to our fastest speed driven example.

Just as is the case for asking Minitab to calculate pooled t-intervals and Welch's t-intervals for \(\mu_1-\mu_2\), the commands necessary for asking Minitab to perform a two-sample t-test or a Welch's t-test depend on whether the data are entered in two columns, or the data are entered in one column with a grouping variable in a second column.

Let's recall the spider and prey example, in which the feeding habits of two species of net-casting spiders were studied. The species, the deinopis, and menneus coexist in eastern Australia. The following data were obtained on the size, in millimeters, of the prey of random samples of the two species:

Let's use the data and Minitab to test whether the mean prey size of the populations of the two types of spiders differs.

1. Enter the data in two columns, such as:

   

2. Under the Stat menu, select Basic Statistics, and then select 2-Sample t...:

   

3. In the pop-up window that appears, select Samples in different columns. Specify the name of the First variable, and specify the name of the Second variable. For the two-sample (pooled) t-test, click on the box labeled Assume equal variances. (For Welch's t-test, leave the box labeled Assume equal variances unchecked.):

   

4. Click on the button labeled Options... In the pop-up window that appears, for the box labeled Alternative, select either less than, greater than, or not equal depending on the direction of the alternative hypothesis:

   

   Then, click OK to return to the main pop-up window.

5. Then, upon clicking OK on the main pop-up window, the output should appear in the Session window:

   Two-Sample T:   For Deinopis vs Menneus

   Variable	N	Mean	StDev	SE Mean
   Deinopis	10	10.26	2.51	0.79
   Menneus	10	9.02	1.90	0.60

   Difference = mu (Deinopis) - mu (Menneus)
   Estimate for difference:  1.240
   95% CI for difference: (-0.852, 3.332)
   T-Test of difference = 0 (vs not =):  T-Value = 1.25   P-Value = 0.229    DF = 18
   Both use Pooled StDev = 2.2266

1. Enter the data in one column (called Prey, say), and the grouping variable in a second column (called Group, say, with 1 denoting a deinopis spider and 2 denoting a menneus spider), such as:

   

2. Under the Stat menu, select Basic Statistics, and then select 2-Sample t...:

   

3. In the pop-up window that appears, select Samples in one column. Specify the name of the Samples variable (Prey, for us) and specify the name of the Subscripts (grouping) variable (Group, for us). For the two-sample (pooled) t-test, click on the box labeled Assume equal variances. (For Welch's t-test, leave the box labeled Assume equal variances unchecked.):

   

4. Click on the button labeled Options... In the pop-up window that appears, for the box labeled Alternative, select either less than, greater than, or not equal depending on the direction of the alternative hypothesis:

   

   Then, click OK to return to the main pop-up window.

5. Then, upon clicking OK on the main pop-up window, the output should appear in the Session window:

   Two-Sample T:   For Prey

   Group	N	Mean	StDev	SE Mean
   1	10	10.26	2.51	0.79
   2	10	9.02	1.90	0.60

   Difference = mu (1) - mu (2)
   Estimate for difference:  1.240
   95% CI for difference: (-0.852, 3.332)
   T-Test of difference = 0 (vs not =):  T-Value = 1.25   P-Value = 0.229    DF = 18
   Both use Pooled StDev = 2.2266