Question

The quality of the drinking water must be monitored as often as possible. One variable of concern is

the pH level, which measures the alkalinity or 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 is targeting higher than 8.0 pH. Based

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

the claim using 5%

Step

1 Describe the population parameter of interest

2 Formulate the null and alternative hypothesis

3 Check the assumptions

4 Choose a signifinance level size for α

5 Select the appropriate test statistic

Compute the test statistic using the appropriate formula

6 State the decision rule for rejecting or not the null hypothesis

7 Compare the computed test statistic and the critical value /s

Answer 1

1. The pH level is the population parameter of interest

2.The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 8$

$H_1:\mu>8$

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

4. Let the significance level be $\alpha = 0.05,$ $df=n-1=15$ and the critical value for a right-tailed test is $t_c = 1.75305.$

The rejection region for this right-tailed test is $R = \{t:t> 1.75305\}.$

5. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{7.6-8}{0.4/\sqrt{16}}=-4$$

Since it is observed that $t=-4< 1.75305=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=15$ degrees of freedom, $t=-4$ is $p=0.99942,$ and since $p=0.99942>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 8, at the $\alpha = 0.05$ significance level.