Question

The Pew Research poll described in Exercise 5 found that 56% of a sample of 1060 teens go online several times a day. (Treat this as a simple random sample.)

Find the margin of error for this poll if we want 95% confi- dence in our estimate of the percent of American teens who go online several times a day.

Answer 1

The margin of error for the Pew Research poll at a 95% confidence level, with a sample proportion of 56% and a sample size of 1060, is approximately 2.9%.

Step-by-step explanation:

To calculate the margin of error for the Pew Research poll with a confidence level of 95%, we can use the formula for the margin of error for a proportion, which is Z * sqrt(p(1-p)/n), where Z is the Z-score corresponding to the confidence level, p is the sample proportion, and n is the sample size.

First, we find the Z-score for a 95% confidence level, which is approximately 1.96. The sample proportion, based on the poll, is 0.56 (56%), and the sample size is 1060 teens. Inserting these values into the formula gives us:Margin of Error = 1.96 * sqrt(0.56(1-0.56)/1060)Calculating this, we find the margin of error to be approximately 0.029 or 2.9%.This means we are 95% confident that the true proportion of American teens who go online several times a day lies within 2.9 percentage points of the observed 56%.