In: Statistics and Probability
A recent survey on the usage of cosmetics among youth provided the following data:
Year: 2004; Sample size: 5000; Youth who use cosmetics: 36%
Year: 2014; Sample size: 4250; Youth who use cosmetics: 47%
Construct a 99% confidence interval for the difference in
population proportions of youth who were using cosmetics in 2004
and youth who were using cosmetics in 2014. Assume that random
samples are obtained and the samples are independent. (Round your
answers to three decimal places.)
z0.10 | z0.05 | z0.025 | z0.01 | z0.005 |
1.282 | 1.645 | 1.960 | 2.326 | 2.576 |
Select the correct answer below:
(−0.150,−0.070)
(−0.136,−0.084)
(−0.140,−0.080)
(−0.127,−0.093)
The confidence interval for difference in proportions is
P1 - P2 +- Z/2 * ((P1 * Q1) / n1 + (P2 * Q2) / n2
Here P1 = 36% = 0.36 youth who use cosmetics in 2004
Q1 = 1 - P1 = 1 - 0.36 = 0.64
n1 = 5000
Here P2 = 47% = 0.47 youth who use cosmetics in 2014
Q2 = 1 - P2 = 1 - 0.47 = 0.53
n2 = 4250
Confidence interval = 99%, So = 0.01 and /2 = 0.005
So Z0.005 = 2.576 as given in the data
SO the confidecne interval for difference in the proportions = P1 - P2 +- Z/2 * ((P1 * Q1) / n1 + (P2 * Q2) / n2 )
= 0.36 - 0.47 +- 2.576 * (0.36 * 0.64) / 5000 + (0.47 * 0.53) / 4250
= -0.11 +- 2.576 * 0.0001
= -0.11 +- 2.576 * 0.01
= -0.11 +-0.0258
(-0.136 , -0.084)
So the Answer is B