Question

You sample a population and find that the standard deviation of the mean in your initial sample is 17. What is the minimum number of observations you need to estimate 90%, 95%, and 99% confidence intervals of the population mean with a margin of error equal to 3?

Answer 1

Answer:

90% confidence

$n=(\frac{1.64(17)}{3})^2 =86.36 \approx 87$

So the answer for this case would be n=87 rounded up to the nearest integer

95% confidence

$n=(\frac{1.960(17)}{3})^2 =123.36 \approx 124$

So the answer for this case would be n=124 rounded up to the nearest integer

99% confidence

$n=(\frac{2.58(17)}{3})^2 =213.744 \approx 214$

So the answer for this case would be n=214 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=17 represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The margin of error is given by this formula:

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

And on this case we have that ME =3 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$   (3)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.64$, replacing into formula (3) we got:

$n=(\frac{1.64(17)}{3})^2 =86.36 \approx 87$

So the answer for this case would be n=87 rounded up to the nearest integer

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$, replacing into formula (3) we got:

$n=(\frac{1.960(17)}{3})^2 =123.36 \approx 124$

So the answer for this case would be n=124 rounded up to the nearest integer

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 (3) we got:

$n=(\frac{2.58(17)}{3})^2 =213.744 \approx 214$

So the answer for this case would be n=214 rounded up to the nearest integer