Question

A 98% confidence interval for the mean of a population is to be constructed and must be accurate to within 0.3 unit. A preliminary sample standard deviation is 1.7. The smallest sample size n that provides the desired accuracy is

Answer 1

Answer:

$n = (\frac{2.326*1.7}{0.3})^2= 173.73$

And the value for n rounded up would be n = 174

Step-by-step explanation:

We have the following info given:

$s= 1.7$ previous estimation for the deviation

$ME=0.03$ the margin of error desired

$Conf =0.98$ represent the confidence

The Margin of error is given by:

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

If we solve for the value of n we got:

$n= (\frac{z*\sigma}{ME})^2$

For this problem we know that the confidence is 98% so then the significance level would be $\alpha=0.02$ and the critical value would be:

$z_{\alpha/2}= 2.326$

And replacing we got:

$n = (\frac{2.326*1.7}{0.3})^2= 173.73$

And the value for n rounded up would be n = 174