Answer:
The 90% confidence interval, C.I = (0.693, 0.7307)
Step-by-step 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, [tex]\overline x_1[/tex] = 85
The sample standard deviation, s₁ = 4
Catalyst 2
The number of batches in the sample, n₂ = 10
The average yield, [tex]\overline x_2[/tex] = 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;
[tex]The \ 90\% \ confidence \ interval= \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2} \times (s_p) \times \sqrt{\dfrac{1^{2}}{n_{1}}+\dfrac{1^{2}}{n_{2}}}[/tex]
[tex]s_p =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}[/tex]
Therefore;
[tex]s_p[/tex] = √((12 - 1)×4² + (10 - 1)×5²)/(12 + 10 - 2)) ≈ 4.48
Therefore, we get;
[tex]C.I. = \left (85- 81 \right )\pm 1.725 \times 4.48 \times \sqrt{\dfrac{1}{12}+\dfrac{1}{10}}[/tex]
The 90% confidence interval, C.I = 0.693 < μ₁ - μ₂ < 7.307 = (0.693, 0.7307).
