90% Confidence Interval for Difference Between Two Proportions

Question

In random, independent samples of 225 adults and 250 teenagers who watched a certain television show, 99 adults and 125 teens indicated that they liked the show. Let $p_1$ be the proportion of all adults watching the show who liked it, and let $p_2$ be the proportion of all teens watching the show who liked it. Find a 90% confidence interval for $p_1-p_2.$

1. what is the lower limit of the 90% confidence interval
2. what is the upper limit of the 90% confidence interval

Answer 1

Here is Minitab output for the given data. You can find the formula in your text and see if your results match results from Minitab. (Negative signs are because I put Adults first so estimated difference is $0.44 - 0.50 = -0.06.$)

Test and CI for Two Proportions 

Sample    X    N  Sample p
1        99  225  0.440000
2       125  250  0.500000

Difference = p (1) - p (2)
Estimate for difference:  -0.06
90% CI for difference:  (-0.135289, 0.0152889)
Test for difference = 0 (vs ≠ 0):  Z = -1.31  P-Value = 0.190

Also, here are my results from R (which uses a continuity correction):

 prop.test(c(99,125), c(225,250), conf=.90)

    2-sample test for equality of proportions with continuity correction

 data:  c(99, 125) out of c(225, 250) 
 X-squared = 1.4785, df = 1, p-value = 0.224
 alternative hypothesis: two.sided 
 90 percent confidence interval:
  -0.13951109  0.01951109 
 sample estimates:
 prop 1 prop 2 
   0.44   0.50 

Answer 2

My first answer was incorrect because I used the wrong proportions, but the method was correct. Since Bruce has the outputs from computer software (which everyone isn't familiar with) then I will supply the standard formula that can be done by hand:

 p1 = .44;
 p2 = .50;
 n1 = 225;
 n2 = 250;

$\left(\text{upper}=0.06\, + 1.64 \sqrt{\frac{\text{p1} (1-\text{p1})}{\text{n1}}+\frac{\text{p2} (1-\text{p2})}{\text{n2}}}\right)$

$\left(\text{lower}=0.06\, -1.64 \sqrt{\frac{\text{p1} (1-\text{p1})}{\text{n1}}+\frac{\text{p2} (1-\text{p2})}{\text{n2}}}\right)$

$\text{upper}=.0.06\, +1.64 \sqrt{\frac{0.44*0.56}{225}+\frac{0.5* 0.5}{250}}$

$\text{lower}=.0.06\, -1.64 \sqrt{\frac{0.44*0.56}{225}+\frac{0.5* 0.5}{250}}$

Thus, the interval is

  (-.015,.135)

Notice the sign difference because I used Teens - Adults. It is the same answer as above, but make sure to interpret it correctly. R is of course a lot easier and supplies much more information