Question

. A survey of randomly chosen adults found that 18% of the 218 women and 22% of the 232 men surveyed had purchased books online. Find a 99% confidence interval for the difference in the proportions of women and men who purchase books online and explain what your interval means.

A) Difference between sample proportions (p̂₁ – p̂₂) =

b) The value of z* =

c) SE of the confidence interval

d) ME of the confidence interval =

e) Confidence interval =

f) Write your explanation of the confidence interval:

Answer 1

a)
Here, , n1 = 218 , n2 = 232
p1cap = 0.18 , p2cap = 0.22

point estimate = 0.18 - 0.22 = -0.04

b)

For 0.99 CI, z-value = 2.58

c)

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.18 * (1-0.18)/218 + 0.22*(1-0.22)/232)
SE = 0.0376

d)

For 0.99 CI, z-value = 2.58

ME = 2.58 * 0.0376 = 0.0970

e)

Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.18 - 0.22 - 2.58*0.0376, 0.18 - 0.22 + 2.58*0.0376)
CI = (-0.137 , 0.057)

f)

we are 95% confident atht the r the difference in the proportions of women and men who purchase books online is between -0.137 and 0.057