Question

The PCB concentration of a fish caught in Lake Michigan was measured by a technique that is known to result in an error of measurement that is normally distributed with a standard deviation of .08 ppm (parts per million). Suppose the results of 10 independent measurements of this fish are 11.2, 12.4, 10.8, 11.6, 12.5, 10.1, 11.0, 12.2, 12.4, 10.6(a) Give a 95 percent confidence interval for the PCB level of this fish.(b) Give a 95 percent lower confidence interval.(c) Give a 95 percent upper confidence interval.

Answer 1

a) The 95% confidence interval would be given by (10.863;12.097)

b)
`(10.980, \infty)`

c) :
`(-\infty,11.980)`

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

Part a) Confidence interval

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

Data: 11.2, 12.4, 10.8, 11.6, 12.5, 10.1, 11.0, 12.2, 12.4, 10.6

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=11.48`

The sample deviation calculated
`s=0.864`

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

`df=n-1=10-1=9`

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(0.025,9)".And we see that
`t_(\alpha/2)=2.26`

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

`11.48-2.26(0.864)/(√(10))=10.863`

`11.48+2.26(0.864)/(√(10))=12.097`

So on this case the 95% confidence interval would be given by (10.863;12.097)

Part b) Give a 95 percent lower confidence interval

On this case we want a interval on this form :
`(\bar X -t_(\alpha,n-1)(s)/(√(n)), \infty)`

So the critical value would be on this case
`t_(\alpha)=1.83` and we can use the following excel code to find it: "=T.INV(1-0.05,9)"

We found the lower limit like this:

`11.48 -1.83(0.864)/(√(10))=10.980`

And the interval would be:
`(10.980, \infty)`

Part c) Give a 95 percent upper confidence interval.

On this case we want a interval on this form :
`(-\infty,\bar X +t_(\alpha,n-1)(s)/(√(n)))`

So the critical value would be on this case
`t_(\alpha)=1.83` and we can use the following excel code to find it: "=T.INV(1-0.05,9)"

We found the lower limit like this:

`11.48+1.83(0.864)/(√(10))=11.980`

And the interval would be:
`(-\infty,11.980)`