Answer:
0.226, 0.374
Explanation:
A high school is running a campaign against the over-use of technology in teens. The committee running the campaign decides to look at the difference in social media usage between teens and adults. and then Find a 90% confidence interval for the difference in proportions.
The formula for confidence interval for the difference between the proportions is given as:
p1 - p2 ± z × √p1 (1 - p1)/n1 + p2(1 - p2)/n2
From the question
We have two groups.
Group 1
They take a random sample of 200 teens in their city (Group 1) and find that 85% of them use social media,
p1 = x/n1
n1 = 200
x1 = 85% × 200 = 170
p1 = 170/200
p1 = 0.85
Group 2
Take another random sample of 180 adults in their city (Group 2) and find that 55% of them use social media.
p2 = x/n1
n2= 180
x2 = 55% × 180 = 99
p2 = 99/180
p2 = 0.55
z = z score for 90% Confidence Interval = 1.645
p1 - p2 ± z × √p1 (1 - p1)/n1 + p2(1 - p2)/n2
= 0.85 - 0.55 ± 1.645 √0.85(1 - 0.85)/200 + 0.55(1 - 0.55)/180
= 0.85 - 0.55 ± 1.645 √0.85(0.15)/200 + 0.55(0.45)/180
= 0.30 ± 1.645 × √0.0020125
= 0.30 ± 1.645 × 0.0448608961
= 0.30 ± 0.0737961741
Hence
= 0.30 - 0.0737961741
= 0.2262038259
= 0.30 + 0.0737961741
= 0.3737961741
Therefore, 90% confidence interval for the difference in proportions is (0.226, 0.374