Question

Suppose a candy company representative claims that colored candies are mixed such that each large production batch has precisely the following proportions: 20% brown, 20% yellow, 10% red, 10% orange, 20% green, and 20% blue. The colors present in a sample of 461 candies was recorded. Is the representative's claim about the expected proportions of each color refuted by the data? Color brown yellow red orange green blue Number of Candies 103 34 115 83 57 69 Step 1. State the null and alternative hypothesis. Step 2. What does the null hypothesis indicate about the proportions of candies of each color? Step 3. State the null and alternative hypothesis in terms of the expected proportions for each category. Step 4. Find the expected value for the number of chocolate candies colored brown. Round your answer to two decimal places. Step 5. Find the expected value for the number of chocolate candies colored orange. Round your answer to two decimal places. Step 6. Find the value of the test statistic. Round your answer to three decimal places. Step 7. Find the degrees of freedom associated with the test statistic for this problem. Step 8. Find the critical value of the test at the 0.01 level of significance. Round your answer to three decimal places. Step 9. Make the decision to reject or fail to reject the null hypothesis at the 0.01 level of significance. Step 10. State the conclusion of the hypothesis test at the 0.01 level of significance.

Answer 1

Solve using Excel:

A B D E (0,-E) / 0.003 0.080 0.223 1 Color 2 Brown 3 Yellow 4. Red 5 Orange 6 Green 7 Blue 8 Total No. of candies Observed Porportion (0) Expected Proportion (E) 103 0.22 0.2 34 0.07 0.2 115 0.25 0.1 83 0.18 0.1 57 0.12 0.2 69 0.15 0.2 461 Test statistic 0.064 0.029 0.013 0.412 9 10 Color 11 Brown 12 Yellow 13 Red 14 Orange 15 Green 16 Blue 17 Total No. of candies Observed Porportion (0) Expected Proportion (E) (0:-)/E 103 =B2/461 0.2 =(C2-D2)^2/02 34 =B3/461 0.2 =(C3-D3)^2/03 115 =B4/461 0.1 =(C4-D4)^2/D4 83 =B5/461 0.1 =(C5-D5)^2/05 57 =B6/461 0.2 =(C6-D6)^2/06 69 =B7/461 0.2 =(C7-D7)^2/07 =SUM(B2:37) Test statistic I=SUM(E2:E7) 18

Color	No. of candies	Observed Porportion (O)	Expected Proportion (E)	(Oi - Ei)2 / Ei
Brown	103	0.22	0.2	0.003
Yellow	34	0.07	0.2	0.080
Red	115	0.25	0.1	0.223
Orange	83	0.18	0.1	0.064
Green	57	0.12	0.2	0.029
Blue	69	0.15	0.2	0.013
Total	461		Test statistic	0.412
Color	No. of candies	Observed Porportion (O)	Expected Proportion (E)	(Oi - Ei)2 / Ei
Brown	103	=B2/461	0.2	=(C2-D2)^2/D2
Yellow	34	=B3/461	0.2	=(C3-D3)^2/D3
Red	115	=B4/461	0.1	=(C4-D4)^2/D4
Orange	83	=B5/461	0.1	=(C5-D5)^2/D5
Green	57	=B6/461	0.2	=(C6-D6)^2/D6
Blue	69	=B7/461	0.2	=(C7-D7)^2/D7
Total	=SUM(B2:B7)		Test statistic	=SUM(E2:E7)

1. H₀: Observed proportion are same as expected proportions

H₁: At least one of the observed proportion is different from expected proportion

2. Null hypothesis indicates that proportion of candies are not the same for each colour.

3. H₀: p₁ = 0.2, p₂ = 0.2, p₃ = 0.1, p₄ = 0.1, p₅ = 0.2, p₆ = 0.2

H₁: Either one of p₁ ≠ 0.2 or p₂ ≠ 0.2 or p₃ ≠ 0.1 or p₄ ≠ 0.1 or p₅ ≠ 0.2 or p₆ ≠ 0.2

4. Expected values for brown candies = 0.2*461 = 92.20

5. Expected values for orange candies = 0.1*461 = 46.10

6. Test statistic: χ² = 0.412

7. Degrees of freedom: df = k-1 = 6-1 = 5

8. Level of significance = 0.01

Critical value (Using Excel CHISQ.INV.RT(probability,df)) = CHISQ.INV.RT(0.01,5) = 15.086

9. Since test statistic (0.412) is less than critical value (15.086), we do not reject the null hypothesis.

10. Observed proportion are same as expected proportions. So, p₁ = 0.2, p₂ = 0.2, p₃ = 0.1, p₄ = 0.1, p₅ = 0.2, p₆ = 0.2.