Question

How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98% confidence that the sample mean is within $400 of the population mean, and the population standard deviation is known to be $1400.

Answer 1

Answer:

$n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67$

So the answer for this case would be n=67 rounded up

Step-by-step explanation:

Information given

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

$\mu$ population mean

$\sigma=1400$ represent the sample standard deviation

n represent the sample size  

Solution to the problem

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 =400 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 98% of confidence interval now can be founded using the normal distribution. And the critical value would be $z_{\alpha/2}=2.33$, replacing into formula (b) we got:

$n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67$

So the answer for this case would be n=67 rounded up