Mathematics, 19.12.2019 20:31, tmrodriguez1

Here are summary statistics for randomly selected weights of newborn girls: n=235, x=30.5 hg, s=6.7 hg. construct a confidence interval estimate of the mean. use a 95% confidence level. are these results very different from the confidence interval 28.9 hg< μ < 31.9 hg with only 12 sample values, x=30.4 hg, and s=2.3 hg?

what is the confidence interval for the population mean μ?

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answered: jeifetz1023

a) The 95% confidence interval would be given by (29.639;31.361 )

And the results are not very different from 28.9 hg< μ < 31.9 hg

b)The 95% confidence interval would be given by (28.939;31.861)

And the results are not very different from 28.9 hg< μ < 31.9 hg

Step-by-step explanation:

1) 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".

$\bar X=30.5$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=6.7 represent the sample standard deviation

n=235 represent the sample size

2) First 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)

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

df=n-1=235-1=234

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,234)".And we see that $t_{\alpha/2}=1.97$

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

$30.5-1.97\frac{6.7}{\sqrt{235}}=29.639$

$30.5+1.97\frac{6.7}{\sqrt{235}}=31.361$

So on this case the 95% confidence interval would be given by (29.639;31.361 )

And the results are not very different from 28.9 hg< μ < 31.9 hg

3) Second confidence interval

$\bar X=30.4$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.3 represent the sample standard deviation

n=12 represent the sample size

df=n-1=12-1=11

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

$30.4-2.20\frac{2.3}{\sqrt{12}}=28.939$

$30.4+2.20\frac{2.3}{\sqrt{12}}=31.861$

So on this case the 95% confidence interval would be given by (28.939;31.861)

And the results are not very different from 28.9 hg< μ < 31.9 hg

answered: Guest

the answer is true.

step-by-step explanation:

answered: Guest

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

A. name the streets that are parallel to golden gate. b. name the streets that intersect 6th street.

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Here are summary statistics for randomly selected weights of newborn girls: n=235, x=30.5 hg, s=6.7...