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

Answer:

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

b)  $(10.980, \infty)$

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

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)

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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:  

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

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

The mean calculated for this case is $\bar X=11.48$

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):

$11.48-2.26\frac{0.864}{\sqrt{10}}=10.863$    

$11.48+2.26\frac{0.864}{\sqrt{10}}=12.097$

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 : $(\bar X -t_{\alpha,n-1}\frac{s}{\sqrt{n}}, \infty)$

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:

$11.48 -1.83\frac{0.864}{\sqrt{10}}=10.980$

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 : $(-\infty,\bar X +t_{\alpha,n-1}\frac{s}{\sqrt{n}})$

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:

$11.48+1.83\frac{0.864}{\sqrt{10}}=11.980$

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