The necessary value for Za/2 was determined to be 1.960. Recall the given information. Sample 1 Sample 2 1 = 400 P₁ = 0.56 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. lower bound = P₁-P₂-²a/2 P₂(1-P₂) 72 = 2 = 300 upper bound = P2 = 0.41 P₁(1-P₁) n1 0.56 0.41 - 1.960 Chill P₁-P₂ + ²a/2V = 0.56 0.41 + 1.960 + Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P₂(1-P₂) P₁(1-P₁) n1 7₂ MT V + )√ 0.56(10.56) 0.41(10.41) 300 400 + 0.56(1-0.56) 0.41(1 0.41) 400 300 Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lower bound of to an upper bound of +
The necessary value for Za/2 was determined to be 1.960. Recall the given information. Sample 1 Sample 2 1 = 400 P₁ = 0.56 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. lower bound = P₁-P₂-²a/2 P₂(1-P₂) 72 = 2 = 300 upper bound = P2 = 0.41 P₁(1-P₁) n1 0.56 0.41 - 1.960 Chill P₁-P₂ + ²a/2V = 0.56 0.41 + 1.960 + Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P₂(1-P₂) P₁(1-P₁) n1 7₂ MT V + )√ 0.56(10.56) 0.41(10.41) 300 400 + 0.56(1-0.56) 0.41(1 0.41) 400 300 Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lower bound of to an upper bound of +
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
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100%

Transcribed Image Text:O
O
The necessary value for Za/2 was determined to be 1.960. Recall the given information.
Sample 1
Sample 2
n₂ = 300
P₂ = 0.41
Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places.
P₁-P2-²a/2
P₂(1-P₂)
7₂
n₁ = 400
P₁ = 0.56
lower bound =
= 0.56 0.41 - 1.960
upper bound =
P₁(1-P₁)
n1
P₁-P₂ + ²a/2V
= 0.56 0.41 +
=
Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places.
P₁(1-P₁) +
P₂(1-P₂)
n₁
7₂
)√
+
+1.960
0.56(1 0.56) 0.41(10.41)
300
✓
400
0.56(1 0.56)
400
+
DELL
+
0.41(10.41)
300
Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lower
to an upper bound of
bound of
Si
74°F Sunny
^ C
40)
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