Mathematics, 17.01.2020 18:31 ahoney2233

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.

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Answer from: natalieburnett82

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

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

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.

Answer from: Quest

idk sry

step-by-step explanation:

Answer from: Quest

Ithink the answer is a

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