Question

Assume that the helium porosity of coal samples taken from any particular seam is Normally distributed with true standard deviation 0.75.a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average porosity of 4.56.
c. How large a sample size is necessary if the width of the 95% interval is to be .40

Answer 1

a)
`4.85-2.33(0.75)/(√(20))=4.46`

`4.85+2.33(0.75)/(√(20))=5.24`

b)
`4.56-2.33(0.75)/(√(16))=4.12`

`4.56+2.33(0.75)/(√(16))=4.99`

c)
`n=((1.960(0.75))/(0.2))^2 =54.02 \approx 55`

Explanation:

Part a

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

The Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that
`z_(\alpha/2)=2.33`

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

`4.85-2.33(0.75)/(√(20))=4.46`

`4.85+2.33(0.75)/(√(20))=5.24`

Part b

`4.56-2.33(0.75)/(√(16))=4.12`

`4.56+2.33(0.75)/(√(16))=4.99`

Part c

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =0.4/2 =0.2 we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got
`z_(\alpha/2)=1.960`, replacing into formula (b) we got:

`n=((1.960(0.75))/(0.2))^2 =54.02 \approx 55`