Question

A soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.1 in. Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 0.25 in.

Answer 1

Answer:

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

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 $z_{\alpha/2}=1.96$, replacing into formula (2) we got:  

$n=(\frac{1.96(0.25)}{0.1})^2 =24.01$  

So the answer for this case would be n=25 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)  

$\sigma=0.25$ represent the population standard deviation

n represent the sample size (variable of interest)  

Solution to the problem

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$  

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 =0.1 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)  

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 $z_{\alpha/2}=1.96$, replacing into formula (2) we got:  

$n=(\frac{1.96(0.25)}{0.1})^2 =24.01$  

So the answer for this case would be n=25 rounded up to the nearest integer