19. In the past, 9% of parts shipped by a vendor have been defective. The vendor claims that they have improved quality, that is, reduced the fraction of defectives in their shipments. To check their claim, you randomly select and test 150 parts from new shipment. You find 12 defectives. Is there evidence, at α = 5%, that the vendor has improved quality?

Question

19.	In the past, 9% of parts shipped by a vendor have been defective.  The vendor claims that they have improved quality, that is, reduced the fraction of defectives in their shipments.  To check their claim, you randomly select and test 150 parts from new shipment.  You find 12 defectives.  Is there evidence, at α = 5%, that the vendor has improved quality?

shiraz14 · Answer

For α = 5% & using a two-tailed probability, 
95% C.I. for p = 0.09 ± (1.96)[sqrt(0.09 x 0.91/150)] = (0.0442, 0.1358)

Given 12 defectives out of 150 parts, p=12/150=0.08. Since this lies within the 95% C.I. computed earlier, there is no significant evidence at α = 5% that the vendor has improved quality.

anonymous · Answer

Where do you get the numbers 1.96 and .91 from?

shiraz14 · Answer

1.96 is the z-value when z=.975 (for a 2-tailed 95% C.I.). You can find this in any table of normal values.
0.91 = 1-0.09 - the formula for a two-tailed C.I. is as written below:

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