Two catalysts in a batch chemical process, are being compared for their effect on the output of the process reaction. A sample o
f 12 batches was prepared using catalyst 1, and a sample of 10 batches was prepared using catalyst 2. The 12 batches for which catalyst 1 was used in the reaction gave an average yield of 85 with a sample standard deviation of 4, and the 10 batches for which catalyst 2 was used gave an average yield 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.
1 answer:
Answer:
2.083 < µ1 - µ2 < 5.917
Step-by-step explanation:
We will need to construct a 90% confidence interval for the difference of 2 means where the populations are normally distributed, and their variances are equal.
The calculations of the sample means and standard deviations are done for us.
Sample 1: Catalyst 1
n = 12, x = 85, s = 4
Sample 2: Catalyst 2
n = 10, x = 81, s = 5
See attached photo for the construction of the confidence interval...
You might be interested in
The height of broken part of tree from ground is 5.569m.
Justification:
Let BD is a tree of height 12 m.
<u>Suppose it got bent at a point C and let the part CD take the position CA, meeting the ground at A</u>.
i.e., CD = AC = h m
<u>Broken part makes 60° angle from ground</u>
So, ∠BAC = 60°
<u>Now, height of remaining part of tree</u> = (12 – h)m.
In right angled ∆ABC,
sin 60° = BC/AC
⇒ √3/2 = (12 - h)/h
⇒ √3h = 2(12 – h)
⇒ √3h = 24 – 2h
⇒ √3h + 2h = 24
⇒ h(√3 + 2) = 24
⇒ h(1.732 + 2) = 24
⇒ h(3.732) = 24
⇒ h = 24/3.732 = 6.4308 m
<u>Hence, height of broken tree from ground</u>
⇒ BC = 12 – h
⇒ 12 – 6.4308 = 5.569m
<u>Hence, tree is broken 5.569 m from ground</u>.
<u>Note</u>: See attached picture.
