Question

In random, independent samples of 200 adults and 400 teenagers who watched a certain television show, 98 adults and 224 teens indicated that they liked the show. Let p1 be the proportion of all adults watching the show who liked it, and let P2 be the proportion of all teens watching the show who liked it. Find a 90% confidence interval for pi-p2. Then complete the table below. carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.) what is the lower limit of the 90% confidence interval? what is the upper limit of the 90% confidence interval?

Answer 1

Solution :

Given that :-

adults indicated that they liked the show : x1 = 98

in random independent sample : n1 = 200

teenagers indicated that they liked the show : x2 = 224

in random independent sample : n2 = 400

Now we need to find out the lower and upper limits of 90% confidence interval

we know that

CI = (P1 - P2) +/- Zsqrt((P1Q1/n1) + (P2Q2/n2))

where,

=> proportion of all adults P1 = x1/n1

= 98/200

= 0.490

Q1 = 1 - P1

= 1- 0.490

= 0.510

=> proportion of all teens P2 = x2/n2

= 224/400

= 0.560

Q2 = 1 - P2

= 1- 0.560

= 0.440

=> For 90% confidence interval , Z = 1.645

=> The 90% confidence interval for p1 - p2 is

=> (P1 - P2) +/- Zsqrt((P1Q1/n1) + (P2Q2/n2))

=> (0.490 - 0.560) +/- 1.645((0.490*0.510/200) + (0.560*0.440/400))

=> -0.07 +/- 1.645 (0.00124) + (0.000616)

=> -0.07 +/- 1.645 (0.00185)

=> -0.07 +/- 0.00304

=> ( -0.07-0.00304)

=> (-0.073)

=>(-0.07 + 0.00304)

=> (-0.066)

$\therefore$ 90% confidence interval of

=> Lower limit =(-0.073)

=> Upper limit =(-0.066)