Question

Note: I've reordered the proportions so that we get a positive (+) value difference.

Given p−1p−1 = 0.90, n₁ = 350 and p⎯⎯2p¯2 = 0.85, n₂ = 400 . Use Table 1.

a.	Construct the 90% confidence interval for the difference between the population proportions. (Round intermediate calculations and final answer to 4 decimal places.)

  Confidence interval is  to .

Answer 1

Solution :

Given that,

= 0.90

1- = 1 - 090 = 0.10

= 0.85

1 - = 1 - 0.85 = 0.15

n₁ = 350

n₂ = 400

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

90% confidence interval for p₁ - p₂ is ,

( - ) - Z_/2  * [(1- ) / n₁ + (1 - ) / n₂] < p₁ - p₂ < ( - ) + Z_/2  * [(1- ) / n₁ + (1 - ) / n₂]

(0.90 - 0.85) - 1.645 * [ (0.90 * 0.10) / 350 +  (0.85 * 0.15) / 400] < p₁ - p₂ <

(0.90 - 0.85) - 1.645 * [ (0.90 * 0.10) / 350 +  (0.85 * 0.15) / 400]

0.05 - 0.0395 < p₁ - p₂ < 0.05 + 0.0395

0.0105 < p₁ - p₂ < 0.0895

99% confidence interval for p₁ - p₂ is : (0.0105 , 0.0895)