Question #113901
2. You are interested in heights of trees on either side of a mountain. You choose a random
sample of ten trees on each side of the mountain and measure their heights in each
sample (Table 1).
Table 1. Heights of trees (m) on the north and south of a mountain
North. South
5.5 3.5
7.0. 3.8
5.0. 4.0
4.0 6.8
6.5 3.3
7.5 6.3
5.8. 6.5
8.0. 7.0
4.0. 6.6
5.4 2.0
Test the data to show whether trees are significantly larger on the northern side of the
mountain at the 5 % significance level. Check that the assumptions are met before choosing
the appropriate test, and give reasons for your choice of test.
1
Expert's answer
2020-05-06T15:31:41-0400

We assume that the height of a tree has normal distribution.

We have two small independent samples (n=10<30,m=10<30)(n=10<30, m=10<30). Population variances are unknown. First we will prove that population variances are equal.

H0:σx2=σy2,H1:σx2σy2α=0.05Using Excel (STDEV.S) we findsx21.887sy23.368We will use the following criterion:F=Sb2Ss2 where Sb2,Ss2 are sample variances of bigger and smallersample respectively.F3.3681.8871.785k1=n1=101=9k2=m1=101=9Frightcr=Fcr(α/2;k1;k2)2.5265.F<Frightcr. So we accept H0.Population variances are equal.Now we can test the hypothesis about populationmeans.H0:x^p=y^p,H1:x^p>yp^α=0.05We will use the following criterion:T=X^Y^(n1)Sx2+(m1)Sy2nm(n+m2)n+mUsing Excel (AVERAGE) we findx^=5.87y^=4.98We have T1.228k=n+m2=18tcr(α;k)1.7341(1.7341,) — critical regionT1.228 is not in the critical region. So we accept H0.At a significance level 0.05 we can say that population means areequal.H_0: \sigma_x^2=\sigma_y^2, H_1: \sigma_x^2\neq\sigma_y^2\\ \alpha=0.05\\ \text{Using Excel (STDEV.S) we find}\\ s_x^2\approx 1.887\\ s_y^2\approx 3.368\\ \text{We will use the following criterion:}\\ F=\frac{S_b^2}{S_s^2}\text{ where } S_b^2, S_s^2\text{ are sample variances of bigger and smaller}\\ \text{sample respectively}.\\ F\approx \frac{3.368}{1.887}\approx 1.785\\ k_1=n-1=10-1=9\\ k_2=m-1=10-1=9\\ F_{right_{cr}}=F_{cr}(\alpha/2;k_1;k_2)\approx 2.5265.\\ F<F_{right_{cr}}.\text{ So we accept } H_0.\\ \text{Population variances are equal.}\\ \text{Now we can test the hypothesis about population}\\ \text{means}.\\ H_0:\hat{x}_p=\hat{y}_p, H_1:\hat{x}_p>\hat{y_p}\\ \alpha=0.05\\ \text{We will use the following criterion:}\\ T=\frac{\hat{X}-\hat{Y}}{\sqrt{(n-1)S_x^2+(m-1)S_y^2}}\sqrt{\frac{nm(n+m-2)}{n+m}}\\ \text{Using Excel (AVERAGE) we find}\\ \hat{x}=5.87\\ \hat{y}=4.98\\ \text{We have } T\approx 1.228\\ k=n+m-2=18\\ t_{cr}(\alpha;k)\approx 1.7341\\ (1.7341,\infty)\text{ --- critical region}\\ T\approx 1.228 \text{ is not in the critical region. So we accept } H_0.\\ \text{At a significance level 0.05 we can say that population means are}\\ \text{equal.}

Trees are not significantly larger on the northern side of the mountain at the 5 % significance level.


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