Question

In: Statistics and Probability

A recent survey on the usage of cosmetics among youth provided the following data: Year: 2004;...

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)

Solutions

Expert Solution

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

orchestra answered 1 year ago

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