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.

Respuesta :

Answer:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (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 [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:  

[tex]n=(\frac{1.96(0.25)}{0.1})^2 =24.01 [/tex]  

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma=0.25[/tex] 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:  

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]  

The margin of error is given by this formula:  

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (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:  

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (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 [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:  

[tex]n=(\frac{1.96(0.25)}{0.1})^2 =24.01 [/tex]  

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

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