Most married couples have two or three personality preferences in common. A rand

Question

Most married couples have two or three personality preferences in common. A random sample of 390 married couples found that 130 had three preferences in common. Another random sample of 576 couples showed that 218 had two personality preferences in common. Let p1 be the population proportion of all married couples who have three personality preferences in common. Let p2 be the population proportion of all married couples who have two personality preferences in common.

(a) Find a 95% confidence interval for p1 – p2. (Use 3 decimal places.)

Explanation / Answer

Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=130
No.Of Observed (n1)=390
P1= X1/n1=0.333
Proportion 2
No. of chances(X2)=218
No.Of Observed (n2)=576
P2= X2/n2=0.378
C.I = (0.333-0.378) ±Z a/2 * Sqrt( (0.333*0.667/390) + (0.378*0.622/576) )
=(0.333-0.378) ± 1.96* Sqrt(0.001)
=-0.045-0.061,-0.045+0.061
=[-0.106,0.016]

Interpretations:
1) We are 95% sure that the interval [-0.106 , 0.016 ] contains the difference between
true population proportion P1-P2
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2
3) Since this Cl does contain a zero we can conclude at true mean
difference is zero      

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