We assume that the height of a tree has normal distribution.
We have two small independent samples "(n=10<30, m=10<30)". Population variances are unknown. First we will prove that population variances are equal.
"H_0: \\sigma_x^2=\\sigma_y^2, H_1: \\sigma_x^2\\neq\\sigma_y^2\\\\\n\\alpha=0.05\\\\\n\\text{Using Excel (STDEV.S) we find}\\\\\ns_x^2\\approx 1.887\\\\\ns_y^2\\approx 3.368\\\\\n\\text{We will use the following criterion:}\\\\\nF=\\frac{S_b^2}{S_s^2}\\text{ where } S_b^2, S_s^2\\text{ are sample variances of bigger and smaller}\\\\\n\\text{sample respectively}.\\\\\nF\\approx \\frac{3.368}{1.887}\\approx 1.785\\\\\nk_1=n-1=10-1=9\\\\\nk_2=m-1=10-1=9\\\\\nF_{right_{cr}}=F_{cr}(\\alpha\/2;k_1;k_2)\\approx 2.5265.\\\\\nF<F_{right_{cr}}.\\text{ So we accept } H_0.\\\\\n\\text{Population variances are equal.}\\\\\n\\text{Now we can test the hypothesis about population}\\\\\n\\text{means}.\\\\\nH_0:\\hat{x}_p=\\hat{y}_p, H_1:\\hat{x}_p>\\hat{y_p}\\\\\n\\alpha=0.05\\\\\n\\text{We will use the following criterion:}\\\\\nT=\\frac{\\hat{X}-\\hat{Y}}{\\sqrt{(n-1)S_x^2+(m-1)S_y^2}}\\sqrt{\\frac{nm(n+m-2)}{n+m}}\\\\\n\\text{Using Excel (AVERAGE) we find}\\\\\n\\hat{x}=5.87\\\\\n\\hat{y}=4.98\\\\\n\\text{We have } T\\approx 1.228\\\\\nk=n+m-2=18\\\\\nt_{cr}(\\alpha;k)\\approx 1.7341\\\\\n(1.7341,\\infty)\\text{ --- critical region}\\\\\nT\\approx 1.228 \\text{ is not in the critical region. So we accept } H_0.\\\\\n\\text{At a significance level 0.05 we can say that population means are}\\\\\n\\text{equal.}"
Trees are not significantly larger on the northern side of the mountain at the 5 % significance level.
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