Question

A researcher wants to determine an interval estimate for the average weight of all gorillas (in pounds). She wants to be 95% certain that she is within 5 pounds of the true average. From past studies, it is known that the standard deviation of the weights of gorillas is 11.6 pounds. How many gorillas should she sample to achieve these results?(a) What is the critical value that corresponds to the given level of confidence? Round to two decimals.(b) What is the minimum sample size necessary to estimate the population mean?

Answer 1

Answer:

a) Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

b) The margin of error is given by this formula:

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

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

And replacing we have:

$n=(\frac{1.960(11.6)}{5})^2 =20.67 \approx 21$

So the answer for this case would be n=21 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".

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

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Part a

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Part b

The margin of error is given by this formula:

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

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

And replacing we have:

$n=(\frac{1.960(11.6)}{5})^2 =20.67 \approx 21$

So the answer for this case would be n=21 rounded up to the nearest integer

Answer 2

The critical value will be 1.96 and the minimum sample size necessary to estimate the population will be 21.

What is a normal distribution?

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

A researcher wants to determine an interval estimate for the average weight of all gorillas (in pounds).

She wants to be 95% certain that she is within 5 pounds of the true average.

From past studies, it is known that the standard deviation of the weights of gorillas is 11.6 pounds.

a)  Since the confidence is 0.95 or 95% then the value of α is 0.05 and α/2 is 0.025 and we can calculate the critical value. Then the value of the z-score will be

$z _{\alpha /2} = 1.96$

b)  The margin of error is given as

$ME = z_{\alpha /2} \times \dfrac{\sigma }{\sqrt{n}}$

We have

Margin of error (ME) = 5

Then we have

$n = (\dfrac{z_{\alpha /2} \times \sigma}{ME})^2\\\\\\n = ( \dfrac{1.96 \times 11.6}{5})^2\\\\n = 20.67 \approx 21$

More about the normal distribution link is given below.

https://brainly.com/question/12421652