Question

Benson Manufacturing is considering ordering electronic components from three different suppliers. The suppliers may differ in terms of quality in that the proportion or percentage of defective components may differ among the suppliers. To evaluate the proportion defective components for the suppliers, Benson has requested a sample shipment of 500 components from each supplier. The number of defective components and the number of good components found in each shipment is as follows.  

	Supplier
Component	A	B	C
Defective	10	30	50
Good	490	470	450

a. Formulate the hypotheses that can be used to test for equal proportions of defective components provided by the three suppliers.

1.	H 0: p 1 ≠ p 2 = p 3
2.	H 0: p 1 = p 2 = p 3
3.	H 0: p 1 ≠ p 2 ≠ p 3

Choose correct answer from above choice  
- Select your answer -123Item 1

H _a:  - Select your answer -All population proportions are equalNot all population proportions are equalItem 2

b. Using a .05 level of significance, conduct the hypothesis test. What is the  p-value? Use Table 12.4.

The  p-value is  - Select your answer -less than or equal .05greater than .05Item 3

What is your conclusion?  
- Select your answer -Conclude that the three suppliers provide equal proportions of defective components.Conclude that the three suppliers do not provide equal proportions of defective components.Item 4

c. Conduct a multiple comparison test to determine if there is an overall best supplier or if one supplier can be eliminated because of poor quality. Round  p _i,  p _j and difference to two decimal places. Round critical value to four decimal places.

Comparison	p i	p j	Difference	n i	n j	Critical Value	Significant Diff > CV
A vs B							- Select your answer -YesNoItem 11
A vs C							- Select your answer -YesNoItem 18
B vs C							- Select your answer -YesNoItem 25

Answer 1

a)

H ₀: p ₁ = p ₂ = p ₃

Ha:Not all population proportions are equal

b)

p value is less than or equal .05

.Conclude that the three suppliers do not provide equal proportions of defective components

c)

p̅1=	0.02
p̅2=	0.06
p̅3=	0.1

for 2 df and 0.05 level: X2 =

5.9915

As Criitcal value =√X2*√(p̅1*(1-p̅1)/n1+p̅2*(1-p̅2)/n2)

						critical val
Comparison	pi	pj	|abs diff|	ni	nj	value	significant diff >CV
1 vs 2	0.02	0.06	0.04	500	500	0.0302	significant difference
1 vs 3	0.02	0.10	0.08	500	500	0.0362	significant difference
2 vs 3	0.06	0.10	0.04	500	500	0.0419	not significant difference