A botanist is interested in comparing the growth response of dwarf pea stems to two different levels of the hormone indoeacetic acid (IAA). Using 16-day-old pea plants, the botanist obtains 5-millimeter sections and floats these sections with different hormone concentrations to observe the effect of the hormone on the growth of the pea stem.

Let X and Y denote, respectively, the independent growths that can be attributed to the hormone during the first 26 hours after sectioning for ( 0.5 ) ( 10 ) −4 ( 0.5 ) ( 10 ) −4 and ( 10 ) −4 ( 10 ) −4 levels of concentration of IAA. The botanist would like to test the null hypothesis H 0 : μ x − μ y =0 H 0 : μ x − μ y =0 against the alternative hypothesis H 1 : μ x − μ y <0 H 1 : μ x − μ y <0 . If we can assume X and Y are independent and normally distributed with common variance, respective random samples of size n and m give a test based on the statistic

T= X ¯ − Y ¯ { [ ( n−1 ) S x 2 +( m−1 ) S Yy 2 ]/( n+m−2 ) }( 1/n+1/m ) = X ¯ − Y ¯ S P 1/n+1/m , T= X ¯ − Y ¯ { [ ( n−1 ) S x 2 +( m−1 ) S Yy 2 ]/( n+m−2 ) }( 1/n+1/m ) = X ¯ − Y ¯ S P 1/n+1/m ,

where S P = ( n−1 ) S X 2 +( m−1 ) S Y 2 n+m−2 . S P = ( n−1 ) S X 2 +( m−1 ) S Y 2 n+m−2 .

T has a t distribution with r=n+m−2 r=n+m−2 degrees of freedom when H 0 H 0 is true and the variances are (approximately) equal. The hypothesis Ho will be rejected in favor of H 1 H 1 if the observed value of T is less than − t α ( n+m−2 ) − t α ( n+m−2 ) .

In the example 1, the botanist measured the growths of pea stem segments, in millimeters, for n=11 observations of X given in the Table 1:

and m=13 observations of Y given in the Table 2:

For these data, x ¯ =1.03, s x 2 =0.24, y ¯ =1.66, x ¯ =1.03, s x 2 =0.24, y ¯ =1.66, and s y 2 =0.35 s y 2 =0.35 . The critical region for testing H 0 : μ x − μ y =0 H 0 : μ x − μ y =0 against H 1 : μ x − μ y <0 H 1 : μ x − μ y <0 is t≤− t 0.05 ( 22 )=−1.717 t≤− t 0.05 ( 22 )=−1.717 . Since H 0 H 0 is clearly rejected at α α =0.05 significance level.