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Step2247 [10]
3 years ago
12

If Upper X overbar equals 62​, Upper S equals 8​, and n equals 36​, and assuming that the population is normally​ distributed, c

onstruct a 99 % confidence interval estimate of the population​ mean, mu.
Mathematics
1 answer:
marishachu [46]3 years ago
3 0

Answer:

The 99% confidence interval would be given by (58.373;65.627)    

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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".

\bar X=62 represent the sample mean

\mu population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size  

Part a: 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=36-1=35

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(0.005,35)".And we see that t_{\alpha/2}=2.72

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

62-2.72\frac{8}{\sqrt{36}}=58.373    

62+2.72\frac{8}{\sqrt{36}}=65.627    

So on this case the 99% confidence interval would be given by (58.373;65.627)    

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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