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. They take a random sample of 200 teens in their city (Group 1) and find that 85% of them use social media, and then take another random sample of 180 adults in their city (Group 2) and find that 55% of them use social media. Find a 90% confidence interval for the difference in proportions.

0.226, 0.374

Step-by-step 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