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

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

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/√(9)}` and the observed value is

`z_(0) = (100-102)/(2/√(9)) = -3`.

(a) The rejection region is given by RR = 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/√(9)}` < 1.6448 when
`\mu = 106`) = P(
`\bar{X}` < 102 + (1.6448)(
`2/√(9)`) when
`\mu = 106`) = P(
`\bar{X}` < 103.0965) when
`\mu = 106`) = P(
`(\bar{X}-106)/(2/√(9)) < (103.0965-106)/(2/√(9))`)) = P(Z < -4.3552) = 6.647274e-06