Answer to Question #347849 in Statistics and Probability for Justine Jose

Question

Drinking water has become an important concern among people. The quality of drinking water

must be monitored as often as possible during the day for possible contamination. Another variable of

concern is the pH level, which measures the alkalinity or the acidity of the water. A pH below 7.0 is

acidic while a pH above 7.0 is alkaline. A pH of 7.0 is neutral. A water-treatment plant has a target pH

of 8.0. Based on 16 random water samples, the mean and the standard deviation were found to be: X¯= 7.6 s = 0.4

Does the sample mean provide enough evidence that it differs significantly from the target

mean? In other words, does the sample come from a population whose mean is the same as the

target pH of ? Use , two-tailed test.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=8.0$

$H_1:\mu\not=8.0$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=15$ and the critical value for a two-tailed test is $t_c =2.131449.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.131449\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{7.6-8.0}{0.4/\sqrt{16}}=-4$$

Since it is observed that $|t|=4>2.131449=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=15$ degrees of freedom, $t=-4$ is $p=0.001159,$ and since $p= 0.001159<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 8.0, at the $\alpha = 0.05$ significance level.