Question

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?

Answer 1

Answer:

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  

$df=n-1=12-1=11$

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(0.025,11)".And we see that $t_{\alpha/2}=2.20$

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