Question

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

a) 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.
b) Explain what that margin of error means.
c) If we only need to be 90% confident, will the margin of error be larger or smaller? Explain.
d) Find that margin of error.
e) In general, if all other aspects of the situation remain the same, would smaller samples produce smaller or larger SEC margins of error?

Answer 1

The margin of error reflects the random sampling error in a poll's result and varies depending on the confidence level and sample size, becoming smaller with higher confidence levels and larger sample sizes.

Step-by-step explanation:

The margin of error for a poll is a statistic expressing the amount of random sampling error in the poll's results. It represents how much the poll's results may vary from the true population value.

To calculate the margin of error for 95% confidence, the standard formula is ME = z*(√(p*(1-p)/n)), where z is the z-score corresponding to the desired confidence level, p is the sample proportion, and n is the sample size.

In this case, with a 56% proportion and a sample size of 1060, the margin of error is smaller than when using a 90% confidence level due to a smaller z-score required.

The margin of error becomes larger with a larger sample size if other factors remain constant due to the relationship between the margin of error and sample size being inversely proportional when the confidence level and population proportion remain constant.