Confidence interval for the following population proportion:
Sample= 1762
Defective disks= 70
Therefore:
[tex]p=\frac{70}{1762}=0.0397[/tex]
And:
[tex]q=(1-p)=1-0.0397=0.9603[/tex]
Where q are the disks that are not defective.
The confidence interval is given by:
[tex]CI=Z_c*\sqrt[\placeholder{⬚}]{\frac{p*q}{n}}[/tex]
Where n is the sample= 1762 and Z_c is the z value for a conficence of 90%, Zc= 1.645 . Replacing:
[tex]CI=1.645*\sqrt[\placeholder{⬚}]{\frac{0.0397*0.9603}{1762}}=0.007652[/tex]
Finally, the intervals are given by:
[tex]p\pm CI[/tex]
Substituing:
[tex]Intervals=0.0397\pm0.007652[/tex]
Answer:
Lower Endpoint:
[tex]0.0397-0.007652=0.032048\approx0.032[/tex]
Upper endpoint:
[tex]0.0397+0.007652=0.047352\approx0.047[/tex]
