1. Estimating Proportions

2. Estimating a Proportion
Suppose we sample 100 circuit boards and find that 8 are defective. We would like to make an inference about the true percentage defective given a sample defective of p = 0.08.

3. Estimating a Proportion
Suppose we sample 100 circuit boards and find that 8 are defective. We would like to make an inference about the true percentage defective given a sample defective of p = 0.08.
Recall that for a large sample (n>30) the binomial may be approximated by the normal distribution. We also know that the mean of the binomial is np and the variance is npq. ) , (
npq np N x
, x = # defects in n items

4. Estimating a proportion) , (
npq np N x n x p = ˆ
Now if
then, or ) , ( ˆ
npq np N p n ) 1 , ( ˆ N
npq np p n -

5. Estimating a Proportion) 1 , ( ˆ N
npq np p n -
Divide through by n and replace pq by gives p ˆ q ˆ ) 1 , ( / ˆ N n q p -

6. Estimating a Proportion) 1 , ( / ˆ N n q p - ) / ˆ ( 1 2 a z n q p P - =

7. Estimating a Proportion) / ˆ ( 1 2 a z n q p P - =
Miracle 21c
occurs n q p z / ˆ 2 a

8. Example
Returning to our circuit board example, suppose a sample of 100 boards yields 8% defective. Compute a 90% confidence interval for the true but unknown proportion defective. n q p z / ˆ 2 a
100 / ) 92 (. 08 .
645 1

9. Example
Returning to our circuit board example, suppose a sample of 100 boards yields 8% defective. Compute a 90% confidence interval for the true but unknown proportion defective.
0.035 < p < 0.125