Question

Use the sample data below to test the hypotheses

H ₀: p ₁ = p ₂ = p ₃

H _a: Not all population proportions are the same

	Populations
Response	1	2	3
Yes	150	150	92
No	100	150	108

where p _i is the population proportion of yes responses for population i. Using a .05 level of significance. Use Table 12.4.

a. Compute the sample proportion for each population. Round your answers to two decimal places.

p̄ ₁ =

p̄ ₂ =

p̄ ₃ =

b. Use the multiple comparison procedure to determine which population proportions differ significantly. Use a .05 level of significance. 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
1 vs 2							- Select your answer -YesNoItem 10
1 vs 3							- Select your answer -YesNoItem 17
2 vs 3

2.

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 15 30 50 Good 485 470 450 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

Q1:

	Population 1	Population 2	Population 3
Yes	150	150	92
No	100	150	108
Total	250	300	200

a) Sample proportion:

p̅₁ = 150/250 = 0.6

p̅₂ = 150/300 = 0.5

p̅₃ = 92/200 = 0.46

b)

df = (r-1)*(c-1) = 2

Critical value of chi square, χ²₀ꓸ₀₅ꓹ₂ = CHISQ.INV.RT(0.05, 2) = 5.991

CV₁₂ = √χ²₀ꓸ₀₅ꓹ₂ *√(p̅₁*(1-p̅₁)/n₁ + p̅₂*(1-p̅₂)/n₂) = √5.991*√(0.6*0.4/250 + 0.5*0.5/300) = 0.1037

CV₁₃ = √χ²₀ꓸ₀₅ꓹ₂ *√(p̅₁*(1-p̅₁)/n₁ + p̅₃*(1-p̅₃)/n₃) = √5.991*√(0.6*0.4/250 + 0.46*0.54/200) = 0.1149

CV₂₃ = √χ²₀ꓸ₀₅ꓹ₂ *√(p̅₂*(1-p̅₂)/n₂ + p̅₃*(1-p̅₃)/n₃) = √5.991*√(0.5*0.5/300 + 0.46*0.54/200) = 0.1115

Comparison	pi	pj	|Absolute diff.|	ni	nj	Critical valur	Significant Diff > CV
1 vs 2	0.6	0.5	0.1	250	300	0.1037	No
1 vs 3	0.6	0.46	0.14	250	200	0.1149	Yes
2 vs 3	0.5	0.46	0.04	300	200	0.1115	No

---------------------

Q2:

	A	B	C
Defective	15	30	50
Good	485	470	450
Total	500	500	500

a) H0: p1 = p2 = p3

Ha: Not all population proportions are the equal

b)

Expected Frequencies
	A	B	C	Total
Defective	500 * 95 / 1500 = 31.6667	500 * 95 / 1500 = 31.6667	500 * 95 / 1500 = 31.6667	95
Good	500 * 1405 / 1500 = 468.3333	500 * 1405 / 1500 = 468.3333	500 * 1405 / 1500 = 468.3333	1405
Total	500	500	500	1500
	(fo-fe)²/fe
Defective	(15 - 31.6667)²/31.6667 = 8.7719	(30 - 31.6667)²/31.6667 = 0.0877	(50 - 31.6667)²/31.6667 = 10.614
Good	(485 - 468.3333)²/468.3333 = 0.5931	(470 - 468.3333)²/468.3333 = 0.0059	(450 - 468.3333)²/468.3333 = 0.7177

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 20.7904

df = (r-1)(c-1) = 2

p-value = CHISQ.DIST.RT(20.7904, 2) = 0.000

The p-value is less than or equal .05

Conclusion:

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

c)

a) Sample proportion:

p̅₁ = 15/500 = 0.03

p̅₂ = 30/500 = 0.06

p̅₃ = 50/500 = 0.1

df = (r-1)*(c-1) = 2

Critical value of chi square, χ²₀ꓸ₀₅ꓹ₂ = CHISQ.INV.RT(0.05, 2) = 5.991

CV₁₂ = √χ²₀ꓸ₀₅ꓹ₂ *√(p̅₁*(1-p̅₁)/n₁ + p̅₂*(1-p̅₂)/n₂) = √5.991*√(0.03*0.97/500 + 0.06*0.94/500) = 0.0320

CV₁₃ = √χ²₀ꓸ₀₅ꓹ₂ *√(p̅₁*(1-p̅₁)/n₁ + p̅₃*(1-p̅₃)/n₃) = √5.991*√(0.03*0.97/500 + 0.1*0.9/500) = 0.0378

CV₂₃ = √χ²₀ꓸ₀₅ꓹ₂ *√(p̅₂*(1-p̅₂)/n₂ + p̅₃*(1-p̅₃)/n₃) = √5.991*√(0.06*0.94/500 + 0.1*0.9/500) = 0.0419

Comparison	pi	pj	|Absolute diff.|	ni	nj	Critical valur	Significant Diff > CV
1 vs 2	0.03	0.06	0.03	500	500	0.0320	No
1 vs 3	0.03	0.1	0.07	500	500	0.0378	Yes
2 vs 3	0.06	0.1	0.04	500	500	0.0419	No