Question

Suppose a candy company representative claims that colored candies are mixed such that each large production batch has precisely the following proportions: 10 % brown, 10 % yellow, 20 % red,10 % orange,30 % green, and 20 % blue. The colors present in a sample of 459 candies was recorded. Is the representative's claim about the expected proportions of each color refuted by the data?

Color number of candies	brown 43	yellow 106	red 89	orange 43	green 112	blue 66

Step 1 of 10: State the null and alternative hypothesis.

Step 2 of 10: What does the null hypothesis indicate about the proportions of candies of each color?

Step 3 of 10: State the null and alternative hypothesis in terms of the expected proportions for each category

Step 4 of 10: Find the expected value for the number of chocolate candies colored brown. Round your answer to two decimal places.

Step 5 of 10: Find the expected value for the number of chocolate candies colored red. Round your answer to two decimal places.

Step 6 of 10: Find the value of the test statistic. Round your answer to three decimal places.

Step 7 of 10: Find the degrees of freedom associated with the test statistic for this problem.

Step 8 of 10: Find the critical value of the test at the 0.10.1 level of significance. Round your answer to three decimal places.

Step 9 of 10: Make the decision to reject or fail to reject the null hypothesis at the 0.10.1 level of significance.

Step 10 of 10: State the conclusion of the hypothesis test at the 0.10.1 level of significance.

Answer 1

Step 1 of 10:  

Null hypothesis Ho:: expected proportion is as per  company representative claims

Alternative hypothesis HA:: expected proportion is not as per  company representative claims

Step 2 of 10: null hypothesis :Ho :pbrown =0.1 , pyellow =0.1 , pred =0.2 ; porange =0.1 ; pgreen=0.3 pblue =0.2

Step 3 of 10:: Alternative hypothesis HA :at least one proportion differs

Step 4 of 10:

expected value =np=459*0.1 =45.9

Step 5 of 10: =459*0.2 =91.8

Step 6 of 10:

applying chi square goodness of fit test:

	relative	observed	Expected		Chi square
category	frequency(p)	Oi	Ei=total*p		R2i=(Oi-Ei)2/Ei
1	0.100	43.0	45.90		0.1832
2	0.100	106.0	45.90		78.6930
3	0.200	89.0	91.80		0.0854
4	0.100	43.0	45.90		0.1832
5	0.300	112.0	137.70		4.7966
6	0.200	66.0	91.80		7.2510
total	1.000	459	459		91.1924
test statistic X2 =		91.192

Step 7 of 10:

degree of freedom =categories-1=

5

Step 8 of 10:

for 0.1 level and 5 df :crtiical value X2 =

9.236

Step 9 of 10: reject Ho

Step 10 of 10:

at least one proportion differs from stated