Question

The following measurements (in picocuries per liter) were recorded by a set of xenon gas detectors installed in a laboratory facility: 476,448.2,431.3 Using these measurements, construct a 99% confidence interval for the mean level of xenon gas present in the facility. Assume the population is approximately normal. Step 4 of 4 : Construct the 99% confidence interval. Round your answer to two decimal places.

Answer 1

`451.833-9.925(22.570)/(√(3))=322.50`

`451.833+9.925(22.570)/(√(3))=581.17`

So on this case the 99% confidence interval would be given by (322.50;581.17)

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)

s represent the sample 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 t_(\alpha/2)(s)/(√(n))` (1)

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

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

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

The mean calculated for this case is
`\bar X=451.833`

The sample deviation calculated
`s=22.570`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=3-1=2`

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(0.005,2)".And we see that
`t_(\alpha/2)=9.925`

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

`451.833-9.925(22.570)/(√(3))=322.50`

`451.833+9.925(22.570)/(√(3))=581.17`

So on this case the 99% confidence interval would be given by (322.50;581.17)