Question

A children's cancer research fund has been set up to collect donations for the

treatment and research of the disease. A random sample of 600 people shows
that 47% of the 300 who were contacted by telephone actually made
contributions, compared with only 34% of the 300 who received email
requests. Which of the formulas calculates the 98% confidence interval for the
difference in the proportions of people who make donations when contacted
by telephone versus those contacted by email?

Answer 1

0.47-0.34 ± 2.33 *√[((0.47(0.53)/300) + ((0.34(0.66)/300)]

Step-by-step explanation:

Given that:

By telephone :

Sample size 1, n1 = 300

Sample size 2, n2 = 300

proportion, p1 = 47% = 0.47 ; 1 - p1 = 1 - 0.47 = 0.53

Proportion, p2 = 34% = 0.34 ; 1 - p2 = 1 - 0.34 = 0.66

Confidence interval, Zcrit at 98% = 2.33

p1 - p2 ± Zcrit*√[((p1(1-p1)/n1) + ((p2(1-p2)/n2)]

0.47-0.34 ± 2.33 *√[((0.47(0.53)/300) + ((0.34(0.66)/300)]