Question

Two injection molding machines are used to make the same part. The part warpage from the two machines have known standard deviations of ?1 = 0.012 and ?2 = 0.018. The engineering team believes that the two machines have the same mean warpages. 12 samples from each machine are collected, and the sample means are calculated at ?̅1 = 0.036 and ?̅2 = 0.034.

(a) Is the engineering team correct? Use the appropriate hypothesis test with α = 0.05 to make this determination.

(b) What is the p-value for this test?

(c) Calculate the 99% CI on the difference in means(are the means the same or not?). State your conclusions.

Answer 1

a) H₀:

    H₁:

The test statistic z = ()/

                              = (0.036 - 0.034)/sqrt((0.012)^2/12 + (0.018)^2/12)

                              = 0.32

At = 0.05, the critical values are z_0.025 = +/- 1.96

Since the the test statistic value is not greater than the positive critical value(0.32 < 1.96), so we should not reject H₀.

At 0.05 significance level, there is sufficient evidence to support the engineering team's claim that the two machines have the same mean warpages.

b) P-value = 2 * P(Z > 0.32)

                 = 2 * (1 - P(Z < 0.32))

                 = 2 * (1 - 0.6255)

                = 0.749

c) At 99% confidence interval the critical value z_0.005 = 2.58

The 99% confidence interval is

() +/- z_0.005 *

= (0.036 - 0.034) +/- 2.58 * sqrt((0.012)^2/12 + (0.018)^2/12)

= 0.002 +/- 0.016

= -0.014, 0.018

Since the interval contains 0, so we can conclude that the means are the same.