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

(1)  

(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 , replacing into formula (2) we got:  

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".  

represent the sample mean for the sample  

population mean (variable of interest)  

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:  

The margin of error is given by this formula:  

(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:  

(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 , replacing into formula (2) we got:  

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

the answer is d. 75%

probably .5

i can on others if necessary.