Question

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

asked Aug 20, 2021 132k views

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Tornseglare · Answer 1 · 2021-08-27T08:41:56+0000

Answer:

n =233

30.3-2.60(7.1)/(√(233))=29.09

30.3+2.60(7.1)/(√(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(2.8)/(√(15))=28.54

30.7+2.98(2.8)/(√(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= (31.51-29.09)/(2)= 1.21

And for the second case we got:

Me= (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

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:

$\bar X \pm t_(\alpha/2)(s)/(√(n))$ (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(7.1)/(√(233))=29.09

30.3+2.60(7.1)/(√(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

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(0.005,14)".And we see that
$t_(\alpha/2)=2.98$