Question

A manufacturer of floor polish conducted a consumer-preference experiment to determine which of five different floor polishes was the most appealing in appearance. A sample of 100 consumers viewed five patches of flooring that had each received one of the five polishes. Each consumer indicated the patch he or she preferred. The lighting and background were approximately the same for all patches. The results are given below. Solve the following using the p-value approach and solve using the classical approach.

Polish	A	B	C	D	E	Total
Frequency	26	15	16	21	22	100

(c) Complete the hypothesis test using α = 0.10.
(i) Find the test statistic. (Round your answer to two decimal places.)


(ii) Find the p-value. (Round your answer to four decimal places.)

Skittles Original Fruit bite-size candies are multicolored candies in a bag, and you can "Taste the Rainbow" with their five colors and flavors: green, lime; purple, grape; yellow, lemon; orange, orange; and red, strawberry. Unlike some of the other multicolored candies available, Skittles claims that their five colors are equally likely. In an attempt to reject this claim, a 4-oz bag of Skittles was purchased and the colors counted. Does this sample contradict Skittle's claim at the .05 level?

Red	Orange	Yellow	Green	Purple
17	19	29	17	17

(a) Find the test statistic. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer bounds exactly.)
______< p < _____

Answer 1

c)

i)

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	0.2000	26.0	20.00		1.800
2	0.2000	15.0	20.00		1.250
3	0.2000	16.0	20.00		0.800
4	0.2000	21.0	20.00		0.050
5	0.2000	22.0	20.00		0.200
total	1.000	100	100		4.1000
test statistic X2 =		4.10

ii) from excel: p value =chidist(4.10,4) = 0.3926

2)

a)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	0.2000	17.0	19.80	-0.63	0.396
2	0.2000	19.0	19.80	-0.18	0.032
3	0.2000	29.0	19.80	2.07	4.275
4	0.2000	17.0	19.80	-0.63	0.396
5	0.2000	17.0	19.80	-0.63	0.396
total	1.000	99	99		5.4949
test statistic X2 =		5.49

ii)

0.2 < p value <0.50

(please try 0.2 < p<0.3 if above comes wrong)