Mathematics, 27.03.2020 01:48 nakeytrag

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?

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Answer from: marinahuntg

n =233

$30.3-2.60\frac{7.1}{\sqrt{233}}=29.09$

$30.3+2.60\frac{7.1}{\sqrt{233}}=31.51$

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

n=15

$30.7-2.98\frac{2.8}{\sqrt{15}}=28.54$

$30.7+2.98\frac{2.8}{\sqrt{15}}=32.85$

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 :

$Me= \frac{31.51-29.09}{2}= 1.21$

And for the second case we got:

$Me= \frac{32.85-28.54}{2}= 2.155$

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

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".

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

$\mu$ 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:

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

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

df=n-1=233-1=232

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 excel command would be: "=-T.INV(0.005,232)".And we see that $t_{\alpha/2}=2.60$

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

$30.3-2.60\frac{7.1}{\sqrt{233}}=29.09$

$30.3+2.60\frac{7.1}{\sqrt{233}}=31.51$

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

For the other info provided we have this:

df=n-1=15-1=14

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

$30.7-2.98\frac{2.8}{\sqrt{15}}=28.54$

$30.7+2.98\frac{2.8}{\sqrt{15}}=32.85$

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 :

$Me= \frac{31.51-29.09}{2}= 1.21$

And for the second case we got:

$Me= \frac{32.85-28.54}{2}= 2.155$

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

Answer from: Quest

Bis the answer! hope this

Answer from: Quest

pretty sure 750 each month!

step-by-step explanation:

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Here are summary statistics for randomly selected weights of newborn girls: nequals233, x overbarequ...