Biological oxygen demand BOD is a pollution index that is mo
Biological oxygen demand (BOD) is a pollution index that is monitored in the treated effluent of paper mills. BOD measures the amount of oxygen required to completely oxidize all the organics in one liter of water. High levels of BOD can be hazardous to aquatic life because the oxygen that would normally be used for their respiration is taken away and used to decompose these organic materials. A certain paper mill has taken 25 samples over the past 3 months. the mean and standard deviation for the sample data are 3.25 and 1.092 ppm respectively. The mill would like to see if this sample indicates that the true average BOD in their treated effluent exceeds the targeted 3.00 ppm. Let = 0.1.
A) The appropriate null/alternative hypothesis pair for this study is:
a. Ho:   3 vs Ha:  > 3
 b. Ho:  < 3 vs Ha:   3     
 c. Ho: x  3 vs Ha: x > 3
 d. Ho:   3 vs Ha:  < 3
 e. Ho:  = 3 vs Ha:   3
 f. Ho:   3 vs Ha:   3
 g. Ho: x = 3 vs x > 3
 h. Ho:  < 3 vs Ha:  > 3
 i. Ho:  = 3 vs Ha:  > 3
 
 
 B) The necessary degrees of freedom for this study is:  
 
 
 C) The test statistic, (ttest), for this data set is:  
      Note: Round your answer to the nearest hundredth.
 
 
 D) The p-value is:  
 
 E) The statistical decision and corresponding English interpretation for this study are:
a. FTR Ho: we cannot conclude that the true BOD in treated effluent exceeds 3 ppm
 b. Reject Ho conclude that the true BOD in treated effluent exceeds 3 ppm     
 c. FTR Ha: conclude that the true BOD in treated effluent exceeds 3 ppm
 d. Reject Ha we cannot conclude that the true BOD in treated effluent exceeds 3 ppm
Solution
Set Up Hypothesis
 Null, H0: U<=3
 Alternate, H1: U>3
 Test Statistic
 Population Mean(U)=3
 Sample X(Mean)=3.25
 Standard Deviation(S.D)=1.092
 Number (n)=25
 we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
 to =3.25-3/(1.092/Sqrt(25))
 to =1.145
 | to | =1.145
 Critical Value
 The Value of |t | with n-1 = 24 d.f is 1.318
 We got |to| =1.145 & | t  | =1.318
 Make Decision
 Hence Value of |to | < | t  | and Here we Do not Reject Ho
 P-Value :Right Tail - Ha : ( P > 1.1447 ) = 0.13181
 Hence Value of P0.1 < 0.13181,Here We Do not Reject Ho
[ANSWERS]
 1.
 a. Ho:   3 vs Ha:  > 3
 2. n-1 = 24
 3. to =1.145 ~ 1.15
 4. Ha : ( P > 1.1447 ) = 0.13181
 5. a. FTR Ho: we cannot conclude that the true BOD in treated effluent exceeds 3 ppm


