Question

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

The 90% confidence interval, C.I = (0.693, 0.7307)

Explanation:

Catalysts 1 and catalyst 2 have the following statistical data;

Catalyst 1

The number of batches in the sample, n₁ = 12

The average yield by the 12 batches of catalyst 1,
`\overline x_1` = 85

The sample standard deviation, s₁ = 4

Catalyst 2

The number of batches in the sample, n₂ = 10

The average yield,
`\overline x_2` = 81

The standard deviation, s₂ = 5

F-test = s₂²/s₁² = 5²/4² = 1.5625

The degrees of freedom, df = n₁ + n₂ - 2

∴ df = 12 + 10 - 2 = 20

The critical-t at 90% confidence level = 1.725

The F-test < The critical-t, we pool the variance

The 90% confidence interval with the assumption of equal variance is given as follows;

`The \ 90\% \ confidence \ interval= \left (\bar{x}_(1)- \bar{x}_(2) \right )\pm t_(\alpha /2) * (s_p) * \sqrt{(1^(2))/(n_(1))+(1^(2))/(n_(2))}`

`s_p =\sqrt{(\left ( n_(1)-1 \right )\cdot s_(1)^(2) +\left ( n_(2)-1 \right )\cdot s_(2)^(2))/(n_(1)+n_(2)-2)}`

Therefore;

`s_p` = √((12 - 1)×4² + (10 - 1)×5²)/(12 + 10 - 2)) ≈ 4.48

Therefore, we get;

`C.I. = \left (85- 81 \right )\pm 1.725 * 4.48 * \sqrt{(1)/(12)+(1)/(10)}`

The 90% confidence interval, C.I = 0.693 < μ₁ - μ₂ < 7.307 = (0.693, 0.7307).