Question

In a survey funded by the University of Washington School of Medicine, 195 of 1,300 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 99% confidence interval estimating the proportion of adult Seattle residents that do not believe they can contract an STI. ROUND ANSWER TO 3 DECIMAL PLACES TO FIND ANSWER!!!!

Preliminary:

1. Is it safe to assume that n≤5% of all residents in Seattle?
   • Yes
   • No
   
   
2. Verify nˆp(1−ˆp)≥10 ROUND ANSWER TO ONE DECIMAL PLACE!
   3. nˆp(1−ˆp)= ROUND ANSWER TO ONE DECIMAL PLACE!

4. Confidence Interval: What is the 99% confidence interval to estimate the population proportion? ROUND ANSWER TO THREE DECIMAL PLACES!

5. ????????? <p< ???

Answer 1

3)

1300 * 0.15 *(1-0.15) = 165.8

4)

sample proportion, = 0.15
sample size, n = 1300
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.15 * (1 - 0.15)/1300) = 0.0099

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0099
ME = 0.0255

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.15 - 2.58 * 0.0099 , 0.15 + 2.58 * 0.0099)
CI = (0.124 , 0.176)

5)

0.124< p< 0.176