Question

A. Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average: tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means.

B. In a hatch chemical process, two catalysts arc being compared for their effect on the output of the process reaction. A sample of 12 batches was prepared using catalyst 1 and a sample of 10 batches was obtained using catalyst 2. The 12 batches for which catalyst 1 was used gave an average yield of 85 with a sample standard deviation of 4, and the second sample gave an average of 81 and a sample standard deviation of 5. Find a 90% confidence interval for the difference between the population means, assuming that the populations are approximately normally distributed with equal variances.

Answer 1

Solution

A . Given,

t(alpha/2 , df) = t(0.05/2 , 50+50-2) = 1.9845

95% CI = (x1 - x2) +/- t*sqrt(s1^2/n1 + s2^2/n2)

substitute values

= (78.3 - 87.2) +/- 1.9845*sqrt(5.6^/50 + 6.3^2/50)

= - 8.9 +/- 2.3656

= (- 11.27 , - 6.53)

B. t(alpha/2 , df) = t(0.1/2, 12+10-2) = 1.7247

90% CI = (x1 - x2) +/- t*sqrt(s1^2/n1 + s2^2/n2)

substitute values

= (85 - 81) +/- 1.7247*sqrt(4^2/12 + 5^2/10)

= 4 +/- 3.3768

= (0.623 , 7.377)