Question

You are interested in estimating the the mean weight of the local adult population of female white-tailed deer (doe). From past data, you estimate that the standard deviation of all adult female white-tailed deer in this region to be 21 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 99% confidence level, to within 6 pounds of the actual weight?

Answer 1

`n=((2.58(21))/(6))^2 =81.54 \approx 82`

So the answer for this case would be n=82 rounded up to the nearest integer

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma=21` represent the estimation for the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =6 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.005;0;1)", and we got
`z_(\alpha/2)=2.58`, replacing into formula (b) we got:

`n=((2.58(21))/(6))^2 =81.54 \approx 82`

So the answer for this case would be n=82 rounded up to the nearest integer

Answer 2

We need a sample of at least 82 female white-tailed deer

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 99% confidence level, to within 6 pounds of the actual weight?

We need a sample of size at least n.

n is found when
`M = 6, \sigma = 21`. So

`M = z*(\sigma)/(√(n))`

`6 = 2.575*(21)/(√(n))`

`6√(n) = 21*2.575`

`√(n) = (21*2.575)/(6)`

`(√(n))^(2) = ((21*2.575)/(6))^(2)`

`n = 81.23`

Rounding up

We need a sample of at least 82 female white-tailed deer