B1.1
MARGIN OF ERROR FOR A POPULATION PROPORTION
The question:
A simple random sample of 100 eighth graders at a large suburban middle school indicated that 83% of them are involved with some type of after-school activity. Find the margin of error and the confidence interval for this sample given that we have 90% confidence in the results.
The question stem: What are they asking you to determine?
Find the margin of error and the confidence interval for this sample given that we have 90% confidence in the results.
What are you given? Notice you are given one sample proportion, 83 out of 100 A simple random sample of 100 eighth graders at a large suburban middle school indicated that 83% of them are involved with some type of after-school activity. Find the margin of error and the confidence interval for this sample given that we have 90% confidence in the results.
Identify your variables:
*simple random sample
n = 100 eighth graders at a large suburban middle school indicated
* = 83% or .83, therefore = 1 - .83 = .17
*90% confidence in the results, which means critical value is 1.645
The formula: for a One proportion z-interval test (1-Prop ZInt) and simply calculate the latter part of the formula.
Question type: One proportion z-interval test
Calculation: Using the T1-83: Use 1-Prop ZInt formula to get the confidence interval and then calculate the mean and the difference from each extreme from the mean which is the margin of error.
Select STAT -> TEST -> scroll to 1- Prop ZInt.
You will need x = .83, n = 100, C-Level: .9, calculate
The Connection!
Notice the x = .83 was obtained by transforming the 83% to .83 and multiplying it times 100 to get . This gives you a confidence interval of (.76821, .89179).
Notice that the connection between the confidence interval and the margin of error is the following: Add the absolute value of each extreme together. In this case the two extremes .76821 + .89179 = 1.66 which when divided by two gives me the mean of .83. Now the connection between the margin of error and .83 is that the margin of error is the difference between the .83 and the extremes.
In this case, .83 - .76821 = .06179 and .89179 - .83 = .06179.
Hence the margin of error is .06179.
Interpretation:
We are 90% confident that the mean proportion of students involved in after school activities is between .76821 and .89179.
Note: You will use the Interval functions in the calculator when you are preparing a confidence interval.
