Let X be the length of needle produced by machine. X follows Normal distribution with standard deviation σ=0.01
The margin of error ME = 0.005 and confidence interval is 95%
We have to find here sample size n from given information
The formula to find sample size when population standard deviation is known is
n =[tex] [\frac{z_{\alpha/2} standard deviation}{ME}] ^{2} [/tex]
where [tex] z_{\alpha/2} [/tex] is critical z value for 95% confidence interval
We have confidence level, c=0.95
α = 1- c = 1-0.95 = 0.05
α/2 = 0.05 /2 = 0.025
z (0.025) is z score value for which probability below -z is 0.025 and above z is 0.025
Using z score table to get z critical value
z = -1.96
For confidence interval calculation we use positive z score value 1.96
The sample size will be then
n = [tex] [\frac{1.96 * 0.01}{0.005}]^{2} [/tex]
n = 15.3664
Rounding sample size to nearest integer n=15
a sample is needed to determine with a precision of ±0.005 inches the mean length of the produced needles to 95% confidence is 15
