Question

Sleep deprivation, CA vs. OR, Part I: According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents.


(a) Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived. (Assume the conditions have already been checked.)(please round to four decimal places)

(b) Determine the margin of error for the 95% confidence interval.


(c) Interpret this interval in the context of the data.

(d) Can we conclude that Oregon residents are more sleep-deprived than California residents? Explain.

Answer 1

a)

Here, , n1 = 11545 , n2 = 4691
p1cap = 0.08 , p2cap = 0.088

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.08 * (1-0.08)/11545 + 0.088*(1-0.088)/4691)
SE = 0.0048

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.08 - 0.088 - 1.96*0.0048, 0.08 - 0.088 + 1.96*0.0048)
CI = (-0.0174 , 0.0014)

b)

z value at 95% = 1.96

MArgin of error = 1.96 * 0.0048
= 0.0094

c)

we are 95% confident taht the difference between the proportions of Californians and Oregonians who are sleep deprived. is between (-0.0174 , 0.0014)

d)
n0, because condfidence interval contains 0