Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed.

How large a sample size is necessary if the width of the 90% interval is to be .20? Sample standard deviation of .65

Answer 1

Solution :

Given that,

standard deviation = =0.65

Margin of error = E = width/2=0.20/2=0.1

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )  

sample size = n = [Z/2* / E] 2

n = ( 1.645 *0.65 /0.1 )2

n =640.47

Sample size = n =641