Respuesta :
Answer:
We need a sample of at least 82 female white-tailed deer
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] 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 [tex]M = 6, \sigma = 21[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]6 = 2.575*\frac{21}{\sqrt{n}}[/tex]
[tex]6\sqrt{n} = 21*2.575[/tex]
[tex]\sqrt{n} = \frac{21*2.575}{6}[/tex]
[tex](\sqrt{n})^{2} = (\frac{21*2.575}{6})^{2}[/tex]
[tex]n = 81.23[/tex]
Rounding up
We need a sample of at least 82 female white-tailed deer
Answer:
[tex]n=(\frac{2.58(21)}{6})^2 =81.54 \approx 82[/tex]
So the answer for this case would be n=82 rounded up to the nearest integer
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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=21[/tex] 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:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (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:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (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 [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.58(21)}{6})^2 =81.54 \approx 82[/tex]
So the answer for this case would be n=82 rounded up to the nearest integer
