Question

Two streams, A and B, suspected of being contaminated, were tested for their degree of acidity. Analysis of the 6 water samples taken from stream A showed that the mean acidity level is 7.52 with a standard deviation of 0.024 while the 5 samples taken at Stream B showed an acidity level of 7.49 with a standard deviation of 0.032. Using a 0.05 significance level, determine whether the streams have different acidity levels.

Answer 1

$H_0: \sigma_A^2=\sigma_B^2, H_1:\sigma_A^2\neq \sigma_B^2\\ \alpha=0.05\\ \text{We will use the following criterion:}\\ F=\frac{S_b^2}{S_{sm}^2}\text{ where } S_b^2, S_{sm}^2\text{ are bigger and smaller sample}\\ \text{variances respectively}.\\ F=\frac{(0.032)^2}{(0.024)^2}\approx 1.78.\\ k_1=5-1=4\\ k_2=6-1=5\\ F_{critical_{right}}=F_{critical}(\alpha/2;k_1;k_2)=F_{critical}(0.025;4;5)\approx 7.3879.\\ F<F_{critical_{right}}.\text{ So we accept } H_0.\\ \text{There is no evidence that population variances are not equal}.$

2 streams:

$A: N_A=6, \overline{x}_A=7.52,s_A=0.024\\ B: N_B=5, \overline{x}_B=7.49, s_B=0.032\\ H_0: \mu_A=\mu_B, H_1: \mu_A\neq\mu_B, \alpha=0.05\\ \text{We assume that the level of acidity has normal distribution}.\\ T=\frac{\overline{X}-\overline{Y}}{\sqrt{(n-1)S_1^2+(m-1)S_2^2}}\sqrt{\frac{nm(n+m-2)}{n+m}}\\ \text{We have } T\approx 1.779.\\ k=n+m-2=6+5-2=9\\ t_{critical}=t_{critical}(\alpha;k)=t_{critical}(0.05;9)\approx 2.2622.\\ (-\infty; -2.2622)\cup (2.2622;\infty)\text{ --- critical region}.\\ T\approx 1.779 \text{ does not fall into the critical region. So we accept } H_0.\\ \text{There is no evidence that streams A and B have different acidity levels}.$