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
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
Get Answers For Free
Most questions answered within 1 hours.