Most water-treatment facilities monitor the quality of their drinking water on an hourly basis. One variable monitored is pH, which measures the degree of alkalinity of acidity in the water. A pH below 7.0 is acidic, one above 7.0 is alkaline, and a pH is 7.0 is neutral. One water-treatment plant has a target pH of 8.5. The mean and standard deviation of 1 hour's test results, based on 17 water samples at this plant, are 8.42 and 0.16, respectively. (a) Set up the null and alternative hypotheses to test whether the mean pH value mu of the water from the plant differs from 8.5. (b) What test statistic should be used based on the available sample information

Answer:

a) The null hypothesis is [tex]H_{0}: \mu = 8.5[/tex] and the alternate hypothesis is [tex]H_{a}: \mu \neq 8.5[/tex]

b) A test statistic of -2.06 should be used.

Step-by-step explanation:

Question a:

One water-treatment plant has a target pH of 8.5.

At the null hypothesis, we test if the mean is close to the target. So, the null hypothesis is:

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

At the alternate hypothesis, we test if it differs from 8.5, so the alternate hypothesis is:

[tex]H_{a}: \mu \neq 8.5[/tex]

Question b:

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

8.5 is tested at the null hypothesis:

This means that [tex]\mu = 8.5[/tex]

The mean and standard deviation of 1 hour's test results, based on 17 water samples at this plant, are 8.42 and 0.16, respectively.

This means that [tex]n = 17, X = 8.42, \sigma = 0.16[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{8.42 - 8.5}{\frac{0.16}{\sqrt{17}}}[/tex]

[tex]z = -2.06[/tex]

A test statistic of -2.06 should be used.

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△RTS ≅ △RVS and △RTU ≅ △STV
△VRS ≅ △VRU and △USR ≅ △USV
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