10.1: Multinomial Experiments

Multinomial experiment

• A probability experiment consisting of a fixed number of trials in which there are more than two possible outcomes for each independent trial.

• A binomial experiment had only two possible outcomes.

• The probability for each outcome is fixed and each outcome is classified into categories.

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Example:• A radio station claims that the distribution of music preferences for

listeners in the broadcast region is as shown below.

Distribution of music Preferences

Classical 4% Oldies 2%

Country 36% Pop 18%

Gospel 11% Rock 29%

Each outcome is classified into categories.

The probability for each possible outcome is fixed.

Chi-Square Goodness-of-Fit Test

Chi-Square Goodness-of-Fit Test

• Used to test whether a frequency distribution fits an expected distribution.

• H0: The frequency distribution FITS the specified distribution.

• Ha: The frequency distribution DOES NOT FIT the specified distribution.

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Example:

• To test the radio station’s claim, the executive can perform a chi-square goodness-of-fit test using the following hypotheses.

H0: The distribution of music preferences in the broadcast region is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock. (claim)

Ha: The distribution of music preferences differs from the claimed or expected distribution.

Chi-Square Goodness-of-Fit Test• Observed frequency O - frequency for the category observed in the sample.

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• Expected frequency E - calculated frequency for the category. Expected frequencies are obtained assuming the specified (or hypothesized)

distribution. The expected frequency for the ith category is: Ei = npi

n = number of trials (sample size) pi = assumed probability of ith category.Example: A marketing executive randomly selects 500 radio music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown at the right. Find the observed frequencies and the expected frequencies for each type of music.

Survey results

(n = 500)Classical 8Country 210Gospel 72Oldies 10Pop 75Rock 125

500(0.04) = 20

500(0.36) = 180

500(0.11) = 55500(0.02) = 10

500(0.18) = 90

500(0.29) = 145

n = 500

Chi-Square Goodness-of-Fit Test

1. The observed frequencies must be obtained by using a random sample.

2. Each expected frequency must be greater than or equal to 5.

If these two conditions are satisfied, then the sampling distribution for the goodness-of-fit test is approximated by a chi-square distribution with k – 1 degrees of freedom, where k is the number of categories and test statistic is:

You may perform a hypothesis test using Table 6 Appendix B to find critical values

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22 ( )O E

E The test is always a right-tailed test.

O = Observed frequency in each categoryE = Expected frequency of each category

Example1: Goodness of Fit TestUse the music preference data to perform a chi-square goodness-of-fit test to test whether the distributions are different. Use α = 0.01.

Survey results (n = 500)

Classical 8Country 210Gospel 72Oldies 10Pop 75Rock 125

Distribution of music preferences

Classical 4%Country 36%Gospel 11%Oldies 2%Pop 18%Rock 29%

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H0: music preference is 4% classical,

36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock

Ha :music preference differs from

the claimed or expected distribution = .01d.f. = n –1 = 6 –1 = 5 01

χ2

0 15.086Type of music

Observed frequency

Expected frequency

Classical 8 20Country 210 180Gospel 72 55Oldies 10 10Pop 75 90Rock 125 145

22 ( )O E

E

2 2 2 2 2 2(8 20) (210 180) (72 55) (10 10) (75 90) (125 145)

20 180 55 10 90 14522.713

Decision: Reject H0There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution.

Example2: Goodness of Fit TestThe manufacturer of M&M’s candies claims that the number of different-colored candies in bags of dark chocolate M&M’s is uniformly distributed. To test this claim, you randomly select a bag that contains 500 dark chocolate M&M’s. The results are shown in the table. Using α = 0.10, perform a chi-square goodness-of-fit test to test the claimed or expected distribution. What can you conclude? (Adapted from Mars Incorporated)

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Color Frequency

Brown 80Yellow 95Red 88Blue 83Orange 76Green 78

d.f. = 6 –1 = 5 n = 500•The claim is that the distribution is uniform, so the expected frequencies of the colors are equal. •To find each expected frequency, divide the sample size by the number of colors.• E = 500/6 ≈ 83.3

H0: distribution of different-colored candies in

bags of dark chocolate M & Ms is uniform.

Ha :distribution of different-colored candies in

bags of dark chocolate M & Ms is not uniform.

0.10

χ20 9.236

Expected Frequency83.383.383.383.383.383.3

22 ( )O E

E

2 2 2 2 2 2(80 83.3) (95 83.3) (88 83.3) (83 83.3) (76 83.3) (78 83.3)

83.3 83.3 83.3 83.3 83.3 83.33.016

Decision: Fail to Reject H0 – there is

not enough evidence to dispute the claim that the distribution is uniform.