Question

A quality control engineer at a manufacturing plant is trying to determine the variation of parts that an operator is making with a particular machine. She takes a sample of nparts and, for each part in the sample, measures how much the part width differs from the specifications for the part. She then uses the mean and standard deviation of these values to estimate the mean value for all parts the operator is making. Constructa 95% confidence interval for the mean value assuming a normal distribution and:n= 25mean of sample = .003 instandard deviation of sample = .001 in

Answer 1

t critical value at 0.05 significance level with 24 df = 2.064

95% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

0.003 - 2.064 * 0.001 / sqrt(25) < < 0.003 + 2.064 * 0.001 / sqrt(25)

0.0026 < < 0.0034

95% C I is ( 0.0026 , 0.0034 )