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 35 samples over the past 3 months. the mean and standard deviation for the sample data are 3.3 and 0.85 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 = 0.025.

A) The appropriate null/alternative hypothesis pair for this study is:

Ho: = 3 vs Ha: 3

Ho: 3 vs Ha: < 3

Ho: = 3 vs Ha: > 3

Ho: x = 3 vs x > 3

Ho: 3 vs Ha: 3

Ho: < 3 vs Ha: > 3

Ho: x 3 vs Ha: x > 3

Ho: 3 vs Ha: > 3

Ho: < 3 vs Ha: 3

B) The necessary degrees of freedom for this study is:  

C) The test statistic, (t_test), for this data set is:

D) The p-value is:  

E) The statistical decision and corresponding English interpretation for this study are:
Reject Ha: we cannot conclude that the true BOD in treated effluent exceeds 3 ppm

FTR Ha: conclude that the true BOD in treated effluent exceeds 3 ppm   

FTR Ho we cannot conclude that the true BOD in treated effluent exceeds 3 ppm

Reject Ho: conclude that the true BOD in treated effluent exceeds 3 ppm

Solution

Set Up Hypothesis
Null, H0: U=3
Alternate, treated effluent exceeds the targeted 3.00 ppm H1: U>3
Test Statistic
Population Mean(U)=3
Sample X(Mean)=3.3
Standard Deviation(S.D)=0.85
Number (n)=35
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =3.3-3/(0.85/Sqrt(34))
to =2.088
| to | =2.088
Critical Value
The Value of |t | with n-1 = 34 d.f is 2.032
We got |to| =2.088 & | t | =2.032
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.088 ) = 0.02218
Hence Value of P0.025 > 0.02218,Here we Reject Ho

ANS:
Ho: = 3 vs Ha: > 3
n-1 = 34
to =2.088
P-Value :Right Tail - Ha : ( P > 2.088 ) = 0.02218
Reject Ho: conclude that the true BOD in treated effluent exceeds 3 ppm