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?

n =233

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

n=15

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 :

And for the second case we got:

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".

represent the sample mean for the sample  

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:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.99 or 99%, the value of and , 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

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

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

For the other info provided we have this:

Since the Confidence is 0.99 or 99%, the value of and , 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

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

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 :

And for the second case we got:

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

step-by-step :

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.