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-BARSIC- [3]
3 years ago
8

Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. What is the 95% confidence in

terval for the population variance?
Mathematics
1 answer:
seraphim [82]3 years ago
3 0

Answer:

13.625-2.365\frac{2.669}{\sqrt{8}}=11.393  

13.625+2.365\frac{2.669}{\sqrt{8}}=15.857  

So on this case the 95% confidence interval would be given by (11.393;15.857)  

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".  

Data: 14, 10, 13, 16, 12, 18, 15, 11

We can calculate the sample mean and deviation with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=13.625 represent the sample mean  

\mu population mean (variable of interest)  

s=2.669 represent the sample standard deviation  

n=8 represent the sample size  

Calculate the 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=8-1=7  

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(0.025,7)".And we see that t_{\alpha/2}=2.365

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

13.625-2.365\frac{2.669}{\sqrt{8}}=11.393  

13.625+2.365\frac{2.669}{\sqrt{8}}=15.857  

So on this case the 95% confidence interval would be given by (11.393;15.857)  

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