Question

Question 5 options: A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds that 77% of them received a passing grade. Create a 99% confidence interval for the population proportion of passing test scores. Enter the lower and upper bounds for the interval in the following boxes, respectively. You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.

Question 6 options: An online retailer wants to estimate the number of visitors that click on their advertisement from a particular website. Of 978 page views in a day, 8% of the users clicked on the advertisement. Create a 90% confidence interval for the population proportion of visitors that click on the advertisement. Enter the lower and upper bounds for the interval in the following boxes, respectively. You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.

Answer 1

#5.
sample proportion, = 0.77
sample size, n = 40
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.77 * (1 - 0.77)/40) = 0.0665

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

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.77 - 2.58 * 0.0665 , 0.77 + 2.58 * 0.0665)
CI = (0.5984 , 0.9416)

#6.
sample proportion, = 0.08
sample size, n = 978
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.08 * (1 - 0.08)/978) = 0.0087

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.08 - 1.64 * 0.0087 , 0.08 + 1.64 * 0.0087)
CI = (0.0657 , 0.0943)