Chebychev's Inequality & Central Limit Theorem

[andrew2786]

Hi,
Here is the problem I have, any help would be greatly appreciated:

In an experiment to determine the mean yield per acre of a new turnip seed, n plots of one acre each are planted with the new seed. Assume that 10cwt per acre is the population standard deviation.

Determine a sample size, n, sufficient to ensure a probability of at least 0.80 that the sample mean (average yield of the sample plots) will not differ from the population mean by more than 2cwt per acre using

a) Chebychev's Inequality and
b) the Central Limit Theorem

Why are the two results different?

Please help, I just don't know how to do this!

Many thanks

Andy

[Rassis]

Hi Andrew,

If I do still remember right, things are like this:

If you assume that the results of the experiment follow a Normal distribution (which is quite likely, having into consideration the Central Limit Theorem), then the number of observations needed is given by:

n = [(Z).(Standard Deviation)/Precision)]^2

Because the Confidence Level is 0.8, then Z = 1.28 and you get: n = (1.28 x 10 / 2)^2 = 41 observations.

If you don’t know what distribution is followed by the observations, you can use the Chebyshev´s inequality instead.

Because the Confidence Level is 0.8, then (1 – 1/c^2) = 0.8. Hence c = 2.236 and you get: n = (2.236 x 10 / 2)^2 = 125 observations.

The answer to your last question is implied in the text above.

Rui