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Irina-Kira [14]
3 years ago
7

I’m just confused help!

Mathematics
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

Step-by-step explanation:

180

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The National Institute of Standards and Technology (NIST) supplies "standard materials" whose physical properties are supposed t
Lady bird [3.3K]

Answer:

10.08-1.64\frac{0.1}{\sqrt{6}}=10.013    

10.08+1.64\frac{0.1}{\sqrt{6}}=10.147    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =0.1 represent the population standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=10.08

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

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

10.08-1.64\frac{0.1}{\sqrt{6}}=10.013    

10.08+1.64\frac{0.1}{\sqrt{6}}=10.147    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

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