Answer to Question #262728 in Statistics and Probability for Rocel

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 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 7.6 and 0.4, respectively. Does the sample mean provide enough evidence that it differs significantly from the target mean? Use 𝛼 = 0.05, two – tailed test.

Answer 1

From the information above, we have,

$\mu=8.0,\space \bar{x}=7.6,\space s=0.4, \space n=16$.

The hypotheses tested are,

$H_0:\mu=8.0$

$Against$

$H_1:\mu\not=8.0$

To perform this test, we will apply the student's t distribution since the population variance is not known as described below,

The test statistic is given as,

$t=(\bar{x}-\mu)/(s/\sqrt{n})$

$t=(7.6-8.0)/(0.4/\sqrt{16})$

$t=-0.4/0.1=-4$

$t$ is compared with the t distribution table value at $\alpha=0.05$ with $(n-1)=16-1=15$ degrees of freedom.

The table value is given as,

$t_{\alpha/2,15}=t_{0.05/2,15}=t_{0.025,15}=2.131$, and the null hypothesis is rejected if $|t|\gt t_{0.025,15}$

Since $|t|=|-4|=4\gt t_{0.025,15}=2.131,$ we reject the null hypothesis and we conclude that the sample mean provide sufficient evidence to show that it differs significantly from the target mean at 5% level of significance.