Question

Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is approximately normal. Construct a 99% confidence interval for the mean magnesium ion concentration.

Answer 1

`163.83-4.03(20.094)/(√(6))=130.77`

`163.83+4.03(20.094)/(√(6))=196.89`

So on this case the 99% confidence interval would be given by (130.77;196.89)

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: 175 177 175 180 138 138

We can calculate the mean and the deviation from these data with the following formulas:

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

`s=\sqrt{(\sum_(i=1)^n (x_i -\bar X)^2)/(n-1)}`

`\bar X=163.83` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=20.093 represent the sample standard deviation

n=6 represent the sample size

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

`df=n-1=6-1=5`

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(0.005,5)".And we see that
`t_(\alpha/2)=4.03`

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

`163.83-4.03(20.094)/(√(6))=130.77`

`163.83+4.03(20.094)/(√(6))=196.89`

So on this case the 99% confidence interval would be given by (130.77;196.89)