Suppose the acceptable level of contaminant in cow\'s milk is 5 units per liter.

Question

Suppose the acceptable level of contaminant in cow's milk is 5 units per liter. Let X denote the level of contaminant in one liter of milk taken from a randomly selected cow. An inspector will take a random sample of 30 cows in your dairy and measure the level of contaminant in one liter of milk taken from each cow. If the mean (bar X) of the sample of n = 30 is bigger than the cut-off value of 5 ppm, the inspector will shut down the dairy. a. (Diary owner's viewpoint) Compute the probability of a forced closing by the inspector, supposing that X follows a distribution with: mu = 4.75 ppm and sigma = 1 ppm? mu = 4.9 ppm and sigma = 1? mu = 4.75 and sigma =2 ppm? mu = 4.9 and sigma = 2 ppm? b. (Inspector's viewpoint) What cut-off value is needed, if the inspector wants to be 95% certain that the dairy is shut down whenever the true mean level of contaminant is 4.5 ppm? Compute this for n = 30, and sigma = 1, then again for n = 30, and sigma = 2. c. Summarize and interpret your findings from (a) and (b).

Explanation / Answer

I (a) Acceptable level of containment in cow's milk = 5 units per liter. Inspector will shut down the dairy if he found mean of the sample above cut off value 5ppm.

so (i) Mean = 4.75 ppm and = 1ppm

so Z = (X - ) / ( /n) = ( 5 - 4.75)/( 1/30) = +1.37

P - value for Z = 1.37 => 0.9147

so P( X > 5; 4.75; 1) = 1- 0.9147 = 0.0853

(ii) Mean = 4.9 ppm and = 1ppm

so Z = (X - ) / ( /n) = ( 5 - 4.9)/( 1/30) = +0.55

P - value for Z = 0.55 =>0.7088

so P( X > 5; 4.9; 1) = 1- 0.7088= 0.2912

(iii) Mean = 4.75 ppm and = 2 ppm

so Z = (X - ) / ( /n) = ( 5 - 4.75)/(2/30) = +0.685

P - value for Z = 0.685 =>0.7088

so P( X > 5; 4.75; 1) = 1- 0.7202= 0.2798

(iv) Mean = 4.9 ppm and = 2 ppm

so Z = (X - ) / ( /n) = ( 5 - 4.9)/(2/30) = +0.274

P - value for Z = 0.274=>0.6080

so P( X > 5; 4.9; 2) = 1- 0.6080= 0.3920

(b) Inspector wants to be 95% certain that dairy should be shut down and true mean level of containment is 4.5 ppm.

To acheive that certainity level P(x> X0 ; 4.5; 1) >= 0.95

so Z - value for P value 95 % so it would be - 1.65

so - 1.65 = (X0 - 4.5) / ( 1/30)

X0 = - 1.65 *  ( 1/30) + 4.5 = 4.20

for n = 30 and = 2 ppm

To acheive that certainity level P(x> X0 ; 4.5; 2) >= 0.95

so Z - value for P value 95 % so it would be - 1.65

so - 1.65 = (X0 - 4.5) / ( 2/30)

X0 = - 1.65 *  ( 2/30) + 4.5 = 3.90

c. we can summarize it from each person point of view. For dairy ownerpoint of view, more sample size of cows and less population standard deviation reduce the chances of forcing down the dairy. But from the point of view of Inspector, reducing sample size increasig population standard deviation will increase its chance to detect contamination level and he can shut off the dairy.

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