Question

The current process has a mean of 2.50 and a std deviation of 0.05. A new process has been suggested by research. What sample size is required to detect a process average shift of 0.02 at the 95% confidence level

Answer 1

Answer:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

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

$n=(\frac{1.96(0.05)}{0.02})^2 =24.01$  

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)  

$\sigma=0.05$ represent the population standard deviation  

n represent the sample size (variable of interest)  

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 =0.02 and we are interested in order to find the value of n, if we solve n from equation (1) we got:  

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

$n=(\frac{1.96(0.05)}{0.02})^2 =24.01$  

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