Question

Applying the Central Limit Theorem:

The amount of contaminants that are allowed in food products is determined by the FDA (Food and Drug Administration). Common contaminants in cow milk include feces, blood, hormones, and antibiotics. Suppose you work for the FDA and are told that the current amount of somatic cells (common name "pus") in 1 cc of cow milk is currently 750,000 (note: this is the actual allowed amount in the US!). You are also told the standard deviation is 122000 cells. The FDA then tasks you with checking to see if this is accurate.

c. What is the standard deviation of the sampling distribution?

d. Assuming that the population mean is 750,000, what is the probability that a simple random sample of 40 1 cc specimens has a mean of at least 776975 pus cells?

e. Is this unusual? Use the rule of thumb that events with probability less than 5% are considered unusual. Yes or No

f. Explain your results above and use them to make an argument that the assumed population mean is incorrect. Structure your essay as follows:

-Describe the population and parameter for this situation.

-Describe the sample and statistic for this situation.

-Give a brief explanation of what a sampling distribution is. Describe the sampling distribution for this situation.

-Explain why the Central Limit Theorem applies in this situation.

-Interpret the answer to part d.

-Use the answer to part e. to argue that the assumed population mean is either correct or incorrect. If incorrect, indicate whether you think the actual population mean is greater or less than the assumed value.

-Explain what the FDA should do with this information.

Answer 1

Solution

Back-up Theory

If a random variable X has mean µ, and variance σ², i.e., standard deviation σ, then

X bar has mean µ, and variance σ²/n and hence standard deviation σ/√n …………………… (1)

CENTRAL LIMIT THEOREM ……………………………………………………………. (2)

Let {X₁, X₂, …, X_n} be a sequence of n independent and identically distributed (i.i.d) random variables drawn from a distribution [i.e., {x₁, x_2, …, x_n} is a random sample of size n] of expected value given by µ and finite variance given by σ². Then, as n gets larger, the distribution of

Z = {√n(Xbar − µ)/σ}, approximates the normal distribution with mean 0 and variance 1 (i.e., Standard Normal Distribution)

i.e., sample average from any distribution follows Normal Distribution with mean µ and variance

σ²/n…………………………………………………………………………………………… (2a)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables …………………………………………………..………………… (3a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) …(3b)

Now to work out the solution,

Let X represent amount of somatic cells (common name "pus") in 1 cc of cow milk

Then µ_X = 750000 and σ_X = 122000 [given] ……………………………………………….. (4)

Given

Part (c)

Vide (1) and (4), standard deviation of sampling distribution is: 122000/√n   Answer 1

Part (d)

If Xbar represents the sample average, then vide (2a) [i.e., CLT], Xbar ~ N(750000, 122000²/n).. (5)

So, given n = 40,

P(Xbar ≥ 776975)

= P[Z ≥ {(776975 – 750000)/(122000/√40)}], where Z ~ N(0, 1) and vide (5).

= P(Z ≥ 1.3984)

= 0.081[using Excel Function: Statistical NORMDIST] Answer 2

By the stipulated thumb-rule in the question, with the above probability, it would seem

the event is not unusual Answer 3

Part (e)

The comment above notwithstanding, 0.081 is still a very low probability and hence it is only fair that the event is treated as unusual. This should mean that had the mean been 750000, it is very unlikely that a sample of 40 1 cc specimens would have a mean of at least 776975 pus cells. This in turn implies that the mean of 750000 is incorrect. Answer 4

Clearly, the mean larger than 750000 Answer 5

Part (f)

Action by FDA

1. If all other considerations permit, revise standard upward.

2. If the above is not feasible, stricter adherence to standard needs be ensured. Answer

DONE