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Most water treatment facilities monitor the quality of their drinking water on an hourly basis. One variable measured is pH, which measures the degree of alkalinity or acidity in the water. A pH below 7.0 is acidic, one above 7.0 is alkaline, and a pH of 7.0 is neutral. One water-treatment plant has a target pH of 8.5. (Most try to maintain a slightly alkaline level) The mean and standard deviation of 1 hour test results based on 49 water samples at this plant are: = 8.42; s = 0.16 Does this sample provide sufficient evidence that the mean pH level in the water differs from 8.5?

Respuesta :

Answer:

There is not enough evidence to conclude that the mean pH level is 8.5.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 8.5

Sample mean, [tex]\bar{x}[/tex] = 8.42

Sample size, n = 49

Alpha, α = 0.05

Sample standard deviation, s = 0.16

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 8.5\\H_A: \mu \neq 8.5[/tex]

We use two-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{8.42 - 8.5}{\frac{0.16}{\sqrt{49}} } = -3.5[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 48 degree of freedom } = \pm 2.01[/tex]

Since, the test statistic does not lies in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

There is not enough evidence to conclude that the mean pH level is 8.5.

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