Question

The mean water temperature downstream from a power plant coolingtower discharge pipe should be no more than 102oF. Pastexperience has indicated that the standard deviation of temperatureis 2oF. The water temperature is measured on 9 randomlychosen days, and the average temperature is found to be100oF.(a) Is there evidence that the water temperature isacceptable atα = 0.05?(b) What is the P-value for thistest?(c) What is the probability of accepting thenull hypothesis at α = 0.05 if the water has atrue mean temperature of 106oF?

Answer 1

Answer:

(a) Yes, there is evidence that the water temperature is acceptable at $\alpha = 0.05$ (b) 0.9987 (c) 6.647274e-06

Step-by-step explanation:

Let X be the random variable that represents the water temperature. The water temperature has been measured on n = 9 randomly chosen days (a small sample), the sample average temperature is $\bar{x}$ = 100°F and $\sigma$ = 2°F. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

$H_{0}: \mu = 102$ vs $H_{1}: \mu > 102$ (upper-tail alternative)

We will use the test statistic

$Z = \frac{\bar{X}-102}{2/\sqrt{9}}$ and the observed value is

$z_{0} = \frac{100-102}{2/\sqrt{9}} = -3$.

(a) The rejection region is given by RR = {z | z > 1.6448} where 1.6448 is the 95th quantile of the standard normal distribution. Because the observed value -3 does not belong to RR, we fail to reject the null hypothesis. In other words,  there is evidence that the water temperature is acceptable at $\alpha = 0.05$.  

(b) The p-value for this test is given by P(Z > -3) = 0.9987

(c) P(Accepting $H_{0}$ when $\mu = 106$) = P(The observed value is not in RR when $\mu = 106$) = P($\frac{\bar{X}-102}{2/\sqrt{9}}$ < 1.6448 when $\mu = 106$) = P($\bar{X}$ < 102 + (1.6448)($2/\sqrt{9}$) when $\mu = 106$) = P($\bar{X}$ < 103.0965) when $\mu = 106$) = P($(\bar{X}-106)/(2/\sqrt{9}) < (103.0965-106)/(2/\sqrt{9})$)) = P(Z < -4.3552) = 6.647274e-06