Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals199​, x overbarequals27.9 ​hg, sequals7.6 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 26.3 hgless thanmuless than30.7 hg with only 20 sample​ values, x overbarequals28.5 ​hg, and sequals3.5 ​hg? What is the confidence interval for the population mean mu​? 26.5 hgless thanmuless than 29.3 hg ​(Round to one decimal place as​ needed.) Are the results between the two confidence intervals very​ different? A. ​Yes, because the confidence interval limits are not similar. B. ​Yes, because one confidence interval does not contain the mean of the other confidence interval. C. ​No, because each confidence interval contains the mean of the other confidence interval. D. ​No, because the confidence interval limits are similar.

Answer 1

Answer:

$27.9-2.60\frac{7.6}{\sqrt{199}}=26.499$  

$27.9+2.60\frac{7.6}{\sqrt{199}}=29.301$  

So on this case the 99% confidence interval would be given by (26.5;29.3)  

$28.5-2.86\frac{3.5}{\sqrt{20}}=26.262$  

$28.5+2.86\frac{3.5}{\sqrt{20}}=30.738$  

So on this case the 99% confidence interval would be given by (26.3;30.7)  

Are the results between the two confidence intervals very​ different?

D. ​No, because the confidence interval limits are similar.

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".  

Using the first info

$\bar X=27.9$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$s=7.6$ represent the population standard deviation  

n=199 represent the sample size  

99% confidence interval  

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

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value.

The degrees of freedom on this case are $df=n-1= 199-1=198$

The excel command would be: "=-T.INV(0.005,198)".And we see that $t_{\alpha/2}=2.60$  

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

$27.9-2.60\frac{7.6}{\sqrt{199}}=26.499$  

$27.9+2.60\frac{7.6}{\sqrt{199}}=29.301$  

So on this case the 99% confidence interval would be given by (26.5;29.3)  

Using the other info

The degrees of freedom on this case are $df=n-1= 20-1=19$

The excel command would be: "=-T.INV(0.005,19)".And we see that $t_{\alpha/2}=2.86$  

$28.5-2.86\frac{3.5}{\sqrt{20}}=26.262$  

$28.5+2.86\frac{3.5}{\sqrt{20}}=30.738$  

So on this case the 99% confidence interval would be given by (26.3;30.7)  

As we can see the two intervals are very similar since the upper and lower limits are similar so then the best answer would be:

Are the results between the two confidence intervals very​ different?

D. ​No, because the confidence interval limits are similar.