Refer to Example 25 in Chapter 1 (see also Example 4 here), and test the null hypothesis

Question

Refer to Example 25 in Chapter 1 (see also Example 4 here), and test the null hypothesis [latex]H_{0}\colon F=G[/latex] at level of significance α = 0.05 by using Theorem 4 and the above corollary.

Accepted Answer

Here [latex]m=n=15,\,{\ z_{0.05}}=1.645,\,{\mathrm{and}}\,z_{0.025=1.96}.[/latex] Then:

[latex]C_{1}\simeq112.50-1.96 imes24.105=112.50-47.2458\simeq65.254,[/latex]

[latex]C_{2}\simeq112.50+47.2458\simeq159.746,[/latex]

[latex]C\ \;\simeq\;112.50-1.645 imes24.105\simeq112.50-39.653=72.847,[/latex]

[latex]C^{\prime}\simeq112.50+39.653=152.153.[/latex]

Next, comparing all 15 × 15 pairs in Table 15.3, we get the observed value of U = 185.
Therefore the hypothesis [latex] extstyle H_{0}[/latex] is rejected in favor of [latex]{H}_{A}^{\prime\prime},[/latex] since U = 185 > 159.746 = [latex]C_{2};[/latex] the null hypothesis is also rejected in favor of [latex]{H}_{A}^{\prime},[/latex] since U = 185 > 152.153 = [latex]C^{\prime}{\mathrm{;}}[/latex] but the null hypothesis is not rejected when the alternative is [latex]H_{A},[/latex] because [latex]U=185 less72.847=C.[/latex]

Question 15.7: Refer to Example 25 in Chapter 1 (see also Example 4 here), ......

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