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 113000 cells. The FDA then tasks you with checking to see if this is accurate.

You collect a random sample of 55 specimens (1 cc each) which results in a sample mean of 773070 pus cells. Use this sample data to create a sampling distribution.

a. Why is the sampling distribution approximately normal?

b. What is the mean of the sampling distribution?


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 55 1 cc specimens has a mean of at least 773070 pus cells?


e. Is this unusual? Use the rule of thumb that events with probability less than 5% are considered unusual.

• No
• Yes

f. Explain your results above and use them to make an argument that the assumed population mean is incorrect. (6 points) Structure your essay as follows:

1. Describe the population and parameter for this situation.
2. Describe the sample and statistic for this situation.
3. Give a brief explanation of what a sampling distribution is.
4. Describe the sampling distribution for this situation.
5. Explain why the Central Limit Theorem applies in this situation.
6. Interpret the answer to part d.
7. 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.
8. Explain what the FDA should do with this information.

Answer 1

a). Because our sample size is greater than 30, the Central limit theorem tells us that the sampling distribution will approximately normally a normal distributed. Because we know the population standard deviation and sample size is large, now we will use normal distribution to find probability.

b). Size,n= 55,. mean of sample xbar= 773070

c). Standard deviation sigma= 113000/55= 2055

d). Probability of sample mean at least 773070 is approximately zero