Respuesta :
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.
