Here are summary statistics for randomly selected weights of newborn girls: nequals233, x overbarequals30.3 hg, sequals7.1 hg. Construct a confidence interval estimate of the mean. Use a 99% confidence level. Are these results very different from the confidence interval 28.5 hgless thanmuless than32.9 hg with only 15 sample values, x overbarequals30.7 hg, and sequals2.8 hg?

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".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=233-1=232[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,232)".And we see that [tex]t_{\alpha/2}=2.60[/tex]

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

[tex]30.3-2.60\frac{7.1}{\sqrt{233}}=29.09[/tex]

[tex]30.3+2.60\frac{7.1}{\sqrt{233}}=31.51[/tex]

So on this case the 99% confidence interval would be given by (29.09;31.51)

For the other info provided we have this:

[tex]df=n-1=15-1=14[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that [tex]t_{\alpha/2}=2.98[/tex]

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

[tex]30.7-2.98\frac{2.8}{\sqrt{15}}=28.54[/tex]

[tex]30.7+2.98\frac{2.8}{\sqrt{15}}=32.85[/tex]

So on this case the 99% confidence interval would be given by (28.54;32.85)

So the margin of error for the first case is :

[tex] Me= \frac{31.51-29.09}{2}= 1.21[/tex]

And for the second case we got:

[tex] Me= \frac{32.85-28.54}{2}= 2.155[/tex]

So we can see that if we increase the sample size the margin of error is significantly lower