Question 15.4: Refer to Example 25 in Chapter 1 regarding the plant height ......
Refer to Example 25 in Chapter 1 regarding the plant height (in 1/8 inches) of cross-fertilized and self-fertilized plants. Denote by X_{i}’s and Y_{i}’s, respectively, the heights of cross-fertilized and self-fertilized plants. Then the observed values for the 15 pairs are given in Example 25 of Chapter 1, which are reproduced in the Table 15.2 for convenience. At the level of significance α = 0.05, test the null hypothesis H_{0}\colon F=G, where F and G are the d.f.’s of the X_{i}’s and Y_{i}’s, respectively.
| Table 15.2 | |||||
| Pair | Cross- | Self- | Pair | Cross- | Self- |
| 1 | 188 | 139 | 9 | 146 | 132 |
| 2 | 96 | 163 | 10 | 173 | 144 |
| 3 | 168 | 160 | 11 | 186 | 130 |
| 4 | 176 | 160 | 12 | 168 | 144 |
| 5 | 153 | 147 | 13 | 177 | 102 |
| 6 | 172 | 149 | 14 | 184 | 124 |
| 7 | 177 | 149 | 15 | 96 | 144 |
| 8 | 163 | 122 | |||
From Table 15.2,we have:
Z_{1}=0,~~Z_{2}=1,~~Z_{3}=0,~~Z_{4}=0,~~Z_{5}=0,~~Z_{6}=0,~~Z_{7}=0,~~Z_{8}=0,
Z_{9}=0,~~Z_{10}=0,~~Z_{11}=0,~~Z_{12}=0,~~Z_{13}=0,~~Z_{14}=0,~~Z_{15}=1,
so that Z = 2. Suppose first that the alternative is H_{A}^{\prime\prime}:\ F\ \neq\ G. Then \textstyle H_{0} is rejected in favor of H_{A}^{\prime\prime} whenever Z\leq C_{1}\;\mathrm{or}\;Z\geq C_{2}, where:
P(Z\lt C_{1})+\gamma_{0}P(Z=C_{1})=0.025\;\mathrm{and}\;P(Z\leq C_{2})-\gamma_{0}P(Z=C_{2})=0.975, and Z\sim B(15,1/2).
From the Binomial tables, we find C_{1}=4,\,C_{2}=11,\mathrm{~and~}\gamma_{0}=\textstyle{\frac{37}{208}}\simeq0.178. Since Z=2\lt C_{1}(=4), the null hypothesis is rejected. Next, test \textstyle H_{0} against the alternative H_{A}^{\prime}\colon p\lt {\textstyle\frac12} again at level α = 0.05. Then \textstyle H_{0} is rejected in favor of H_{A}^{\prime} whenever Z\leq C^{\prime},{\mathrm{where}}\;C^{\prime} is determined by:
P(Z\lt C^{\prime})+\gamma^{\prime}P(Z=C^{\prime})=0.05,\quad Z\sim B(15,~1/2).
From the Binomial tables, we find C^{\prime}\,=\,4\,\operatorname{and}\,\gamma^{\prime}\,=\,{\textstyle\frac{81}{104}}\,\simeq\,0.779. Since Z=2\lt C^{\prime}(=4),\,H_{0} is rejected in favor of H_{A}^{\prime}, which is consistent with what the data say.
For the Normal approximation, we get from (15):
\left.\begin{array}{l}{{~C\simeq{\frac{n}{2}}+z_{\alpha}{\frac{\sqrt{n}}{2}}},~~~C^{\prime}\simeq{\frac{n}{2}}-z_{\alpha}{\frac{\sqrt{n}}{2}}}\\ {{C_{1}\simeq{\frac{n}{2}}-z_{\frac{\alpha}{2}}\frac{\sqrt{n}}{2},~~~C_2\simeq\frac{n}{2}+z_{\frac{\alpha}{2}}\frac{\sqrt{n}}{2}.}}\end{array}\right\} (15)
{\bf z}_{0.025}\:=\:1.96, so that C_{1}=3.703,C_{2}\simeq11.297, and \textstyle H_{0} is rejected again, since Z=2\lt C_{1}(\simeq3.703). Also, z_{0.05}={1}.645, and hence C^{\prime}\simeq4.314.. Again, \textstyle H_{0} is rejected in favor of H_{A}^{\prime}, since Z=2\lt C^{\prime}(\simeq4.314).