Question

A coworker claims that Skittles candy contains equal quantities of each color (purple, green, orange, yellow, and red). In other words, 1/5 of all Skittles are purple, 1/5 of all Skittles are green, etc. You, an avid consumer of Skittles, disagree with her claim. Test your coworker's claim at the α=0.01α=0.01 level of significance, using the data shown below from a random sample of 200 Skittles.

Which would be correct hypotheses for this test?

• H0:H0: Red Skittles are cherry flavored; H1:H1: Red Skittles are strawberry flavored
• H0:H0:Skittles candy colors come in equal quantities; H1:H1:Skittles candy colors do not come in equal quantities
• H0:H0:Taste the Rainbow; H1:H1:Do not Taste the Rainbow
• H0:p1=p2H0:p1=p2; H1:p1≠p2H1:p1≠p2

Sample Skittles data:

Color	Count
Purple	49
Green	49
Orange	43
Yellow	31
Red	28

Test Statistic:



Give the P-value. Express to two decimal places:



Which is the correct result:

• Do not Reject the Null Hypothesis
• Reject the Null Hypothesis

Which would be the appropriate conclusion?

• There is not enough evidence to reject the claim that Skittles colors come in equal quantities.
• There is not enough evidence to support the claim that Skittles colors come in equal quantities

Answer 1

Solution:

Given:

Claim: Skittles candy contains equal quantities of each color (purple, green, orange, yellow, and red).

Part a) Which would be correct hypotheses for this test?

H0: Skittles candy colors come in equal quantities;

H1: Skittles candy colors do not come in equal quantities

Part b) Test Statistic:

Chi square test statistic for goodness of fit

AN (=^

Where

O_i = Observed Counts

E_i =Expected Counts = N / k = 200 /5 = 40

Thus we need to make following table

Color	Oi: Observed Count	Ei: Expected counts	Oi2/Ei
Purple	49	40	60.025
Green	49	40	60.025
Orange	43	40	46.225
Yellow	31	40	24.025
Red	28	40	19.600
	N = 200		= 209.900

Thus

AN (=^

X2 = 209.900 – 200

y = 9.900

Part c) Give the P-value. Express to two decimal places:

To get exact P-value , use Excel command:

=CHISQ.DIST.RT( x , df )

where

x = and df = k - 1 = 5 -1 = 4

y = 9.900

=CHISQ.DIST.RT(9.900,4)

=0.0421

=0.04

P-value = 0.04

To get P-value for Chi-square table, look in table for df = 4 row and find interval in which 9.990 fall and then find corresponding right tail area.

X2990 X.100 X1050 X2025 X2995 0.000 0.010 0.072 020 0.000 0.020 0.115 0297 0.001 0.051 0216 0.484 X2350 0.004 0.103 0.352 x2000 0.016 0.211 0.584 LC4 2.706 4.605 6.251 3541 5.991 7.15 5024 7 378 9.348 11.143 9.488 1

fall between 9.488 and 11.143

y = 9.900

thus 0.025 < P-value < 0.050

Part d)   the correct result:

Since P-value = 0.04 > 0.01 level of significance, we  Do not Reject the Null Hypothesis.

Part e) the appropriate conclusion:

There is not enough evidence to reject the claim that Skittles colors come in equal quantities.