Question

(1 point) Suppose we want a 95% confidence interval for the average amount spent on books by freshmen in their first year at college. The amount spent has a normal distribution with standard deviation $16. (a) How large should the sample be if the margin of error is to be less than $5? ANSWER: (b) If we wanted a smaller margin of error, we would need a sample. (Enter: ''LARGER'', ''SMALLER'' or ''SAME SIZE'', without the quotes.)

Answer 1

Answer:

a) $n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

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

b) For this case if we see the formula for the margin of error

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

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

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

$\sigma=16$ represent the population standard deviation

n represent the sample size  

Solution to the problem

Part a

The margin of error is given by this formula:

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

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

$n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

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

Part b

For this case if we see the formula for the margin of error

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

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER