Researchers are interested if a school breakfast program leads to taller children. Assume that the population of all 5 year-old heights are normally distributed with a mean (mu) of 38 inches and a standard deviation (sigma) of 1 inch. The 25 children in the school breakfast program have a mean height of 39 inches. Compute the 95% confidence interval for the population mean if all children were in the school breakfast program. Hints: The sample mean should be used in the formula for the 95% confidence interval. Compute the standard error first, then the margin of error.

So on this case the 95% confidence interval would be given by (38.608;39.392)    

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=25 represent the sample size  

Solution to the problem

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

  (1)

The margin of error is given by:

Since the Confidence is 0.95 or 95%, the value of and , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that

And replacing we got:

Now we have everything in order to replace into formula (1):

So on this case the 95% confidence interval would be given by (38.608;39.392)    

could you provide figure.