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5. A random sample of parents from a population was questioned in 2000 and a second (independent) random sample from the same population was questioned in 2015. Each respondent was asked whether or not they condone the slapping of young children as a regular form of punishment. Of 1460 parents questioned in 2000, 841 condoned slapping and of 1347 questioned in 2015, 364 condoned slapping. (a) Construct a 99% confidence interval for the difference in the true proportion of parents who condone slapping in 2015 compared to 2000. (b) Interpret the confidence interva

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Answer #1

(a)

From given information:

hat{p}_{1}=rac{364}{1347}=0.2702

841 1460 P2 =-= 0.5760

Here we have x1-364, n-1347,P. =0.2702 x2-841,n2-1460, P2 -0.576 The standard error is: SE-ja(1-pl) +P2 (1-p2) =0.0177 1 72 Level of significance: α=0.01 Critical value: Z2.576 The requried confidence interval is: p - p2)+ Z, SE (0.2702 0.576).576*0.0177) --0.3058+0.0456 =(-0.3514,-0.2602) Excel function for critical value NORMSINV(1-(0.01/2)) Hence, the required confidence interval is (-0.3514,-0.2602)

(b)

We are 99% confident that true difference in proportion of parents who condone slapping in 2015 compared to 2000 lie in the above interval.

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