It is desired to estimate the mean GPA of each undergraduate class at a large university. How large a sample is necessary to estimate the GPA within at the confidence level? The population standard deviation is . If needed, round your final answer up to the next whole number.

So the answer for this case would be n=154 rounded up to the nearest integer

Step-by-step explanation:

Assuming the following question: It is desired to estimate the mean GPA of each undergraduate class at a large university. How large a sample is necessary to estimate the GPA within 0.25 at the 99% confidence level? The population standard deviation is 1.2. If needed, round your final answer up to the next whole number.

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)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

   (a)

And on this case we have that ME =0.25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=154 rounded up to the nearest integer

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step-by-step explanation:

answer:

c 60% on edg