4.2 Test of the equality of two independent normal distributions

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 $\left(0.5\right){\left(10\right)}^{-4}$ and ${\left(10\right)}^{-4}$ levels of concentration of IAA. The botanist would like to test the null hypothesis $H_0:{\mu }_x-{\mu }_y=0$ against the alternative hypothesis $H_1:{\mu }_x-{\mu }_y<0$ . 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

$$T=\frac{\overline{X}-\overline{Y}}{\sqrt{\left\{\left[\left(n-1\right)S_x^2+\left(m-1\right)S_{Yy}^2\right]/\left(n+m-2\right)\right\}\left(1/n+1/m\right)}}=\frac{\overline{X}-\overline{Y}}{S_P\sqrt{1/n+1/m}},$$

where $$S_P=\sqrt{\frac{\left(n-1\right)S_X^2+\left(m-1\right)S_Y^2}{n+m-2}}.$$

T has a t distribution with $r=n+m-2$ degrees of freedom when ${\text{H}}_{\text{0}}$ is true and the variances are (approximately) equal. The hypothesis Ho will be rejected in favor of ${\text{H}}_{\text{1}}$ if the observed value of T is less than $-t_{\alpha }\left(n+m-2\right)$ .

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, $\overline{x}=1.03,s_x^2=0.24,\overline{y}=1.66,$ and $s_y^2=0.35$ . The critical region for testing $H_0:{\mu }_x-{\mu }_y=0$ against $H_1:{\mu }_x-{\mu }_y<0$ is $t\le -t_{0.05}\left(22\right)=-1.717$ . Since ${\text{H}}_{\text{0}}$ is clearly rejected at $\alpha$ =0.05 significance level.