Question

Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year.

Answer 1

Answer:

$n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689$

So the answer for this case would be n=689 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: "Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.

Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year."

Solution for the problem

First we need ti find the estimation for the standard deviation using the Rule of thumb, with the following formula:

$s \approx \frac{R}{4}$

Where R is the range defined as :

$R = Max - Min = 8-0 = 8$

So then the deviation would be approximately:

$s \approx 4$

Important 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 (ME) 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".

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 $ME =\pm 0.25$ 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)

We can assume that the estimator for the population deviation from the rule of thumb is $\hat \sigma = s= 4$

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

$n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689$

So the answer for this case would be n=689 rounded up to the nearest integer