Question

3. A survey of 100 Californians finds reports that 48% are excited by the opportunity to take a statistics class.

(a) Construct a 95% confidence interval on the true proportion of Californians who are excited to take a statistics class.

(b) If the sample size is 500, construct a 95% confidence interval on the population proportion.

(c) If we increase the sample size to 1000, construct a 95% confidence interval on the population proportion.

(d) Can you explain briefly what happens to the margin of error and the confidence interval as that sample size increases?

Answer 1

a)

sample proportion, = 0.48
sample size, n = 100
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.48 * (1 - 0.48)/100) = 0.05

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.05
ME = 0.098

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.48 - 1.96 * 0.05 , 0.48 + 1.96 * 0.05)
CI = (0.38 , 0.58)

b)

sample proportion, = 0.48
sample size, n = 500
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.48 * (1 - 0.48)/500) = 0.0223

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0223
ME = 0.0437

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.48 - 1.96 * 0.0223 , 0.48 + 1.96 * 0.0223)
CI = (0.44 , 0.52)

c)
sample proportion, = 0.48
sample size, n = 1000
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.48 * (1 - 0.48)/1000) = 0.0158

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0158
ME = 0.031

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.48 - 1.96 * 0.0158 , 0.48 + 1.96 * 0.0158)
CI = (0.45 , 0.51)

d)
As the sample size increases the margin of error and confidence interval decreases