Question

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.

Answer 1

Answer:

$39-1.96\frac{1}{\sqrt{25}}=38.608$    

$39+1.96\frac{1}{\sqrt{25}}=39.392$    

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".

$\bar X=39$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=1$ 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:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The margin of error is given by:

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, 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 $z_{\alpha/2}=1.96$

And replacing we got:

$ME= 1.96 *\frac{1}{\sqrt{25}}= 0.392$

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

$39-1.96\frac{1}{\sqrt{25}}=38.608$    

$39+1.96\frac{1}{\sqrt{25}}=39.392$    

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