Question

According to recent data by the American Marketing Association, 77.5% of people in the 18-24 age group used social media in 2017. You want to find out what the social media proportion is among your classmates who are all between 18 and 24 years old Suppose you obtain a random sample of 249 students and find that 254 students in this sample are active social media users Complete parts a through c below

a. Construct a 99% confidence interval to estimate the actual proportion of social media users
A99% confidence interval to estimate the actual proportion has a lower limit of_____ and an upper limit of ___________(Round to three decimal places as needed)

Answer 1

To construct a 99% confidence interval to estimate the actual proportion of social media users among your clasportion of social media userssmates, calculate the sample proportion, find the z-score corresponding to the desired confidence level, calculate the margin of error, and construct the confidence interval. So, a 99% confidence interval to estimate the actual proportion has a lower limit of 0.98 and an upper limit of 1.062.

Step-by-step explanation:

To construct a 99% confidence interval to estimate the actual proportion of social media users among your classmates, we can use the formula: CI = p ± z * √((p(1-p))/n), where p is the sample proportion, z is the z-score corresponding to the desired confidence level, and n is the sample size.

Step 1:

Calculate the sample proportion: p = 254/249 = 1.021

Step 2:

Calculate the z-score corresponding to a 99% confidence level. Looking up the z-score in a table, the z-score for a 99% confidence level is approximately 2.576

Step 3:

Calculate the margin of error: ME = z * √((p(1-p))/n) = 2.576 * √((1.021(1-1.021))/249) = 0.041

Step 4:

Construct the confidence interval: CI = p ± ME = 1.021 ± 0.041

Therefore, the 99% confidence interval to estimate the actual proportion of social media users among your classmates is: [0.98, 1.062]