Researchers are interested in determining whether more men than women prefer Coca Cola to Pepsi. In a random sample of 300 men, 65% prefer Coca Cola, whereas in a random sample of 400 women, 48% prefer Coca Cola. What is the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi? 0.17 ± 0.036 0.17 ± 0.096 0.17 ± 0.067 0.565 ± 0.067 0.565 ± 0.096

Respuesta :

Answer:

0.17±0.096

Step-by-step explanation:

p1=0.65

p2=0.48

The 99% confidence interval can be calculated as

[tex](p1-p2)-Z_{\frac{\alpha }{2} } \sqrt{\frac{p1q1}{n1} +{\frac{p2q2}{n2}} }<P1-P2<(p1-p2)+Z_{\frac{\alpha }{2} } \sqrt{\frac{p1q1}{n1} +{\frac{p2q2}{n2}} }[/tex]

q1=1-p1=1-0.65=0.35

q2=1-p2=1-0.48=0.52

n1=300

n2=400

[tex]Z_{\frac{\alpha }{2} } =Z_{\frac{0.01 }{2} }=Z_{0.005} =2.58[/tex]

[tex](0.65-0.48)-2.58 \sqrt{\frac{0.65(0.35)}{300} +{\frac{0.48(0.52)}{400}} }<P1-P2<(0.65-0.48)+2.58 \sqrt{\frac{0.65(0.35)}{300} +{\frac{0.48(0.52)}{400}} }[/tex][tex]0.17-2.58 \sqrt{\frac{0.2275}{300} +{\frac{0.2496}{400}} }<P1-P2<0.17+2.58 \sqrt{\frac{0.2275}{300} +{\frac{0.2496}{400}} }[/tex]

[tex]0.17-2.58 \sqrt{0.000758 +{0.000624} }<P1-P2<0.17+2.58 \sqrt{0.000758 +{0.000624} }[/tex]

[tex]0.17-2.58 \sqrt{0.001382 }<P1-P2<0.17+2.58 \sqrt{0.001382 }[/tex]

[tex]0.17-2.58 (0.037175)<P1-P2<0.17+2.58 (0.037175)[/tex]

[tex]0.17-0.0959<P1-P2<0.17+0.0959[/tex]

0.17±0.096

Thus, the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi is 0.17±0.096