Question

An industrial company claims that the mean PH level of the water in a nearby river is 6.8. You randomly selected 16 water samples and measure the PH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company's claim at 05? Assume the population is normally distributed. Set up appropriate hypotheses, calculate your test statistic, find the rejection region or P-value and make the final conclusion.

Answer 1

Given that, sample size ( n ) = 16

sample mean $(\bar x)$ = 6.7

sample standard deviation ( s ) = 0.24

significance level $(\alpha)$ = 0.05

The null and the alternative hypotheses are,

$H_0: \mu = 6.8$

$H_a: \mu \neq 6.8$

Test statistic is,

$t = \frac { \bar x -\mu}{\frac { s }{\sqrt { n}}}= \frac { 6.7-6.8}{\frac { 0.24}{\sqrt { 16}}} = -1.667$

t-critical value at significance level $(\alpha)$ = 0.05 with degrees of freedom = n - 1 = 16 - 1 = 15 are, t^* = +/- 2.131

Rejection Region : Reject the null hypothesis ( H₀), if test statistic < -2.131 OR test statistic > +2.131

P-value = 0.1163

Here, test statistic = -1.667 > -2.131

so, we fail to reject the null hypothesis ( H₀).

Conclusion: there is not sufficient evidence to warrant rejection of the claim that the mean PH level of the water in a nearby river is 6.8