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

`27.9-2.60(7.6)/(√(199))=26.499`

`27.9+2.60(7.6)/(√(199))=29.301`

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

`28.5-2.86(3.5)/(√(20))=26.262`

`28.5+2.86(3.5)/(√(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.

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)(s)/(√(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(7.6)/(√(199))=26.499`

`27.9+2.60(7.6)/(√(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(3.5)/(√(20))=26.262`

`28.5+2.86(3.5)/(√(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.