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 19 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 98% confidence level, to within 7 pounds of the actual weight?

Answer 1

Answer:

$n=(\frac{2.33(19)}{7})^2 =39.996 \approx 40$

Step-by-step 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  

$\mu$ population mean

$\sigma=19$ represent the assumed population standard deviation

n represent the sample size (variable of interest)  

Confidence =0.98 or 98%

Me=0.07 represent the margin of error for this case

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$  (1)  

And on this case we have that ME =7 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (2)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01,0,1)", and we got $z_{\alpha/2}=2.33$, replacing into formula (2) we got:

$n=(\frac{2.33(19)}{7})^2 =39.996 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer