Question

Two machines are supposed to be producing steel bars of approximately the same length.  A sample of 35 bars from one machine has an average length of 37.013 inches, with a standard deviation of 0.095 inches.  For 38 bars produced by the other machine, the corresponding figures are 36.074 inches and 0.032 inches.  Using a 0.05 level of significance and assuming equal population standard deviations, can we conclude that the machines are not different from each other?  Construct and interpret the 95 percent confidence interval for the difference between the population means

Answer 1

H₀:

H₁:

The pooled variance (s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2) = (34 * (0.095)^2 + 37 * (0.032)^2)/(35 + 38 - 2) = 0.005

The test statistic t = ()/sqrt(s_p²/n1 + s_p²/n2)

= (37.013 - 36.074)/sqrt(0.005/35 + 0.005/38)

= 56.682

At 0.05 significance level the critical values are

+/- t_0.025 = +/- 1.994

Since the test statistic value is greater than the positive critical value (56.682 > 1.994), so we should reject the null hypothesis.

At 0.05 significance level we cannot conclude that the machines are not different from each other.

b) The 95% confidence interval for difference in population means is

() +/- t^* sqrt(s_p²/n1 + s_p²/n2)

= (37.013 - 36.074) +/- 1.994 * sqrt(0.005/35 + 0.005/38)

= 0.939 +/- 0.033

= 0.906, 0.972

We are 95% confidence that the difference between the true population means lies between the interval bounds 0.906 and 0.972.