Respuesta :
Answer:
No, there is not enough evidence to reject the company’s claim at the α = 0.05 level of significance.
Step-by-step explanation:
We are given that an industrial company claims that the mean pH level of the water in a nearby river is 6.8.
You randomly select 29 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.35, respectively.
Let [tex]\mu[/tex] = mean pH level of the water in a nearby river.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 6.8 Â Â {means that the mean pH level of the water in a nearby river is 6.8}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu\neq[/tex] 6.8 Â Â {means that the mean pH level of the water in a nearby river is different from 6.8}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
             T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex]Â = sample mean pH level = 6.7
       s = sample standard deviation = 0.35
       n = sample of water = 29
So, test statistics  =  [tex]\frac{6.7-6.8}{\frac{0.35}{\sqrt{29} } }[/tex]   ~ [tex]t_2_8[/tex]
                =  -1.539
Now at 0.05 significance level, the t table gives critical values of -2.048 and 2.048 at 28 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that the mean pH level of the water in a nearby river is 6.8 which means the company’s claim was correct.
