Question

A coworker claims that Skittles candy contains equal quantities of each color (purple, green, orange, yellow,...

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
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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

Oi = Observed Counts

Ei =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 = y = 9.900 and df = k - 1 = 5 -1 = 4

=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 x20

y = 9.900 fall between 9.488 and 11.143

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.

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