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levacccp [35]
3 years ago
14

Here are summary statistics for randomly selected weights of newborn​ girls: nequals=174174​, x overbarxequals=30.930.9 ​hg, seq

uals=7.57.5 hg. Construct a confidence interval estimate of the mean. Use a 9595​% confidence level. Are these results very different from the confidence interval 29.629.6 hgless than
Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0

Answer:

The 95% confidence interval would be given by (29.780;32.020)  

Step-by-step explanation:

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=30.9 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size  

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=174-1=173

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that t_{\alpha/2}=1.97, this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.  

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

30.9-1.97\frac{7.5}{\sqrt{174}}=29.780    

30.9+1.97\frac{7.5}{\sqrt{174}}=32.020

So on this case the 95% confidence interval would be given by (29.780;32.020)    

The value 29.6 is not contained on the interval calculated.

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