A94 Two identical pencil making machines are side by side As

A.9.4. Two identical pencil making machines are side by side. Assume the lengths Xi of pencils made by machine 1 are normally distributed with mean mu 1 (inches) and variance sigma^2 = .0004 (inches^2) and the lengths Yj of pencils made by machine 2 are also normally distributed with mean mu^2 and variance sigma^2 = .0004. If a random sample of size 18 from machine 1 produced a sample mean of bar x = 8.007 and a random sample of size 15 from machine 2 produced a sample mean of bar y = 7.998. (a) Construct a 95% confidence interval for difference in mean pencil lengths mu1 - mu2. (b) Construct a 99% confidence interval for difference in mean pencil lengths mu1 - mu2. What differs in these two constructions?

Solution

(a) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

(xbar-ybar) - Z*sqrt(s1^2/n1+s2^2/n2)

=(8.007-7.998) - 1.96*sqrt(0.0004/18+0.0004/15)

=-0.004704436

So the upper bound is

(xbar-ybar) + Z*sqrt(s1^2/n1+s2^2/n2)

=(8.007-7.998) + 1.96*sqrt(0.0004/18+0.0004/15)

=0.02270444

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(b) Given a=1-0.99=0.01, Z(0.005) = 2.58 (from standard normal table)

So the lower bound is

(xbar-ybar) - Z*sqrt(s1^2/n1+s2^2/n2)

=(8.007-7.998) - 2.58*sqrt(0.0004/18+0.0004/15)

=-0.009039512

So the upper bound is

(xbar-ybar) + Z*sqrt(s1^2/n1+s2^2/n2)

=(8.007-7.998) + 2.58*sqrt(0.0004/18+0.0004/15)

=0.02703951

The difference is the confidence interval level