Metin E. answered • 03/17/24
Experienced Community College Teacher Specializing in Statistics
Setup
We are trying to build a confidence interval for a parameter p
which is a population proportion and can be defined as
p = the proportion out of all American adults who used the Internet within the past month to obtain medical information.
A study is conducted on a sample of 1,019 randomly chosen adults
so the sample size is n = 1,019
In the sample, 60% had used the Internet within the past month to obtain medical information.
So the sample proportion, p-hat, is given by p-hat = 60% = 0.6
Part 1.
We want to create a 95% confidence interval for the population proportion based on the given information.
The sample was randomly chosen, which is a required condition.
The sample size is large enough to use the Normal distribution in building this confidence interval.
n * p-hat = 1,019 * 0.6 = 611.4
n * (1 - p-hat) = 1,019 * (1 - 0.6) = 1,019 * 0.4 = 407.6
Depending on your textbook, these values might be compared to 10 or 15 as a benchmark.
They are an order of magnitude greater than any benchmark value so we are good.
The critical value from a Normal distribution corresponding to a 2-sided 95% confidence interval is approximately z = 1.96.
Why? Because in a Normal distribution, about 95% of the values are between -1.96 and 1.96.
The formula for a confidence interval for a population proportion is given by:
p-hat ± z * (p-hat * (1 - p-hat) / n)^0.5
So, here we get:
0.6 ± 1.96 * (0.6 * 0.4 / 1,019)^0.5
0.6 - 1.96 * (0.6 * 0.4 / 1,019)^0.5 ≈ 0.5699
0.6 + 1.96 * (0.6 * 0.4 / 1,019)^0.5 ≈ 0.6301
The 95% confidence interval estimate for the percentage of all American adults who have used the Internet to obtain medical information in the past month is given by (0.5699, 0.6301) or (56.99%, 63.01%).
Part 2.
The critical value from a Normal distribution corresponding to a 2-sided 90% confidence interval is approximately z = 1.645.
Why? Because in a Normal distribution, about 90% of the values are between -1.645 and 1.645.
The formula for a confidence interval for a population proportion is given by:
p-hat ± z * (p-hat * (1 - p-hat) / n)^0.5
So, here we get:
0.6 ± 1.645 * (0.6 * 0.4 / 1,019)^0.5
0.6 - 1.645 * (0.6 * 0.4 / 1,019)^0.5 ≈ 0.5748
0.6 + 1.645 * (0.6 * 0.4 / 1,019)^0.5 ≈ 0.6252
The 90% confidence interval estimate for the percentage of all American adults who have used the Internet to obtain medical information in the past month is given by (0.5748, 0.6252) or (57.48%, 62.52%).
Part 3.
The margin of error for part 1. (95% confidence interval) was 1.96 * (0.6 * 0.4 / 1,019)^0.5 ≈ 0.0301
The margin of error for part 2. (90% confidence interval) was 1.645 * (0.6 * 0.4 / 1,019)^0.5 ≈ 0.0252
The 90% confidence interval has a smaller margin of error than the 95% confidence interval.
Think about it this way: the true population proportion is an invisible fish which you are trying to catch with the help of your nets, the confidence intervals.
The larger your net, the higher your chances of catching the fish.
So the net that gives you a 90% chance of catching the fish is smaller than the one that gives you a 95% chance of catching the fish.
Part 4.
The 95% confidence interval is more likely to contain the true percentage of all American adults who have used the Internet to obtain medical information in the past month.
It is a wider net, so it is more likely to have caught the fish.
