Question

Consolidated Power, a large electric power utility, has just built a modern nuclear power plant. This plant discharges waste water that is allowed to flow into the Atlantic Ocean. The Environmental Protection Agency (EPA) has ordered that the waste water may not be excessively warm so that thermal pollution of the marine environment near the plant can be avoided. Because of this order, the waste water is allowed to cool in specially constructed ponds and is then released into the ocean. This cooling system works properly if the mean temperature of waste water discharged is 60°F or cooler. Consolidated Power is required to monitor the temperature of the waste water. A sample of 100 temperature readings will be obtained each day, and if the sample results cast a substantial amount of doubt on the hypothesis that the cooling system is working properly (the mean temperature of waste water discharged is 60°F or cooler), then the plant must be shut down and appropriate actions must be taken to correct the problem.

(a) Consolidated Power wishes to set up a hypothesis test so that the power plant will be shut down when the null hypothesis is rejected. Set up the null hypothesis H0 and the alternative hypothesis Ha that should be used. H0: μ (Click to select)>≤ 60 versus Ha: μ (Click to select)≤> 60.

(b) Suppose that Consolidated Power decides to use a level of significance of α = .05 and suppose a random sample of 100 temperature readings is obtained. If the sample mean of the 100 temperature readings is formula310.mmlx ¯ x¯ = 60.726, test H0 versus Ha and determine whether the power plant should be shut down and the cooling system repaired. Perform the hypothesis test by using a critical value and a p-value. Assume σ = 4. (Round your z to 2 decimal places and p-value to 4 decimal places.)

z

p-value

(Click to select)Reject or Do not reject H0. So the plant (Click to select)should or should not be shut down and the cooling system repaired.

Answer 1

a) H0: < 60

H₁: > 60

The test statistic z = ()/()

= (60.726 - 60)/(4/)

= 1.82

P-value = P(Z > 1.82)

= 1 - P(Z < 1.82)

= 1 - 0.9656 = 0.0344

Since the p-value is less than the significance level (0.0344 < 0.05), so we should reject the null hypothesis.

Reject H₀. So the plant should be shutdown and the cooling system repaired