A1.1
CONFIDENCE INTERVAL 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 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 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 confidence interval for this sample given that we have 90% confidence in the results.
*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)
Question type: One proportion z-interval test
Calculation: Using the T1-83: Use 1-Prop ZInt
Select STAT -> TEST -> scroll to 1- Prop ZInt.
Identify your variables:
You will need x = 83, n = 100, C-Level: .9, calculate
Notice the x = 83 was obtained by transforming the 83% to .83 and multiplying it by 100. Once you input these values into the TI-83 you obtain the confidence interval of (.76821, .89179).
The variations to 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.
➢ In a survey 83 out of 100 eighth graders at a large suburban middle school indicated they 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.
➢ You could have also been given raw data in which case you would have had 83 pieces of information to place into list 1 and then run the same 1-prop ZInt test in your calculator.
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.
