Answer to Question #115269 in Statistics and Probability for Jennie Campos

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

T-test for two Means – Unknown Population Standard Deviations

The provided sample means are : $\bar{X}_1=7.52,\bar{X}_2=7.48.$

The provided sample standard deviations are: $s_1=0.024,s_2=0.032,$

and the sample sizes are $n_1=6,n_2=5.$

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 $df=9.$ 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 $t_c=2.262,$ for $\alpha=0.05$ and $df=9.$ 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:

$$t={\bar{X}_1-\bar{X}_2\over \sqrt{{(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}({1\over n_1}+{1\over n_2})}}=$$

$$={7.52-7.48\over \sqrt{{(6-1)(0.024)^2+(5-1)(0.032)^2\over 6+5-2}({1\over 6}+{1\over 5})}}\approx1.78$$

Since it is observed that $|t|=1.78\leq2.262,$ 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.