Question

The mean water temperature downstream from a discharge pipe at a power plant cooling tower should be no more than 100°F. Past studies has shown that the standard deviation of temperature is 2°F. The water temperature was measured on 10 randomly chosen days, and the results are shown below: 96.15 97.81 97.09 95.08 99.46 100.03 99.74 98.85 95.08 98.71 a. State the null hypothesis and alternative hypothesis for the above problem. b. Is there evidence that the water temperature is acceptable at α = 1%? c. Compute the 95% confidence interval for the mean temperature of water.

Answer 1

a)

Ho :   µ ≥ 100  
Ha :   µ <   100   (Left tail test)

b)

Level of Significance ,    α =    0.010  
population std dev ,    σ =    2.0000  
Sample Size ,   n =    10  
Sample Mean,    x̅ = ΣX/n =    97.8000  
          
'   '   '  
          
Standard Error , SE = σ/√n =   2/√10=   0.6325  
Z-test statistic= (x̅ - µ )/SE =    (97.8-100)/0.6325=   -3.4785  
          
critical z value, z* =       -2.3263   [Excel formula =NORMSINV(α/no. of tails) ]
          
p-Value   =   0.0003   [ Excel formula =NORMSDIST(z) ]
Decision:   p-value≤α, Reject null hypothesis       
Conclusion: There is enough evidence that   the water temperature is acceptable

c)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.960   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   2/√10=   0.6325          
margin of error, E=Z*SE =   1.9600   *   0.6325   =   1.2396
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    97.80   -   1.2396   =   96.5604
Interval Upper Limit = x̅ + E =    97.80   -   1.2396   =   99.0396
95%   confidence interval is (   96.56   < µ <   99.04   )