Question

Problem:
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

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df=5+6-2.$ In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is $t_c=2.262, \alpha=0.05, df=9.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.262\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$${\bar{X_1}-\bar{X_2}\over \sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}=$$

$$={7.52-7.49\over \sqrt{\dfrac{(6-1)0.024^2+(5-1)0.032^2}{6+5-2}(\dfrac{1}{6}+\dfrac{1}{5})}}\approx$$

$$\approx1.78$$

Since it is observed that $|t|=1.78\leq2.262=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.1089,$ and since $p=0.1089\geq0.05,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the 0.05 significance level.