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

Respuesta :

Answer:

a) [tex]4.85-2.33\frac{0.75}{\sqrt{20}}=4.46[/tex]    

[tex]4.85+2.33\frac{0.75}{\sqrt{20}}=5.24[/tex]    

b) [tex]4.56-2.33\frac{0.75}{\sqrt{16}}=4.12[/tex]    

[tex]4.56+2.33\frac{0.75}{\sqrt{16}}=4.99[/tex]  

c) [tex]n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55[/tex]

Step-by-step explanation:

Part a

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

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

The Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], 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 [tex]z_{\alpha/2}=2.33[/tex]

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

[tex]4.85-2.33\frac{0.75}{\sqrt{20}}=4.46[/tex]    

[tex]4.85+2.33\frac{0.75}{\sqrt{20}}=5.24[/tex]    

Part b

[tex]4.56-2.33\frac{0.75}{\sqrt{16}}=4.12[/tex]    

[tex]4.56+2.33\frac{0.75}{\sqrt{16}}=4.99[/tex]  

Part c  

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (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:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (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 [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55[/tex]

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