Question

In: Statistics and Probability

A multinomial experiment produced the following results: (You may find it useful to reference the appropriate...

A multinomial experiment produced the following results: (You may find it useful to reference the appropriate table: chi-square table or F table)

A multinomial experiment produced the following results: (You may find it useful to reference the appropriate table: chi-square table or F table)

Category 1 2 3 4 5
Frequency 64 37 67 59 57

a. Choose the appropriate alternative hypothesis to test if the population proportions differ.

  • All population proportions differ from 0.20.

  • Not all population proportions are equal to 0.20.

b. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

c. Find the p-value.

  • 0.025 p-value < 0.05
  • 0.01 p-value < 0.025
  • p-value < 0.01

  • p-value 0.10
  • 0.05 p-value < 0.10

d. Can we conclude at the 10% significance level that the population proportions are not equal?

  • No, since the p-value is less than the significance level.

  • Yes, since the p-value is less than the significance level.

  • No, since the p-value is more than the significance level.

  • Yes, since the p-value is more than the significance level.

Solutions

Expert Solution

Answer: A multinomial experiment produced the following results:

Solution:

Proportion p = 1/5 = 0.20

a) The null and alternative hypothesis test:

Ho: p1 = p2 = p3 = p4 = p5 = 0.20

Ha: Not all population proportions are equal to 0.20.

b) From the given data we used chi square test statistic χ2 at α = 0.05 significance level.

n = 64+37+67+59+57

n = 284

Expected frequency = n*p

Expected frequency = 284 * 0.20

Expected frequency = 56.8

Category Observed (O) Expected (E) (O-E)^2/E
1 64 56.8 0.9127
2 37 56.8 6.9021
3 67 56.8 1.8317
4 59 56.8 0.0852
5 57 56.8 0.0007
Total 284 284 9.7324

Test statistic χ2 = Σ(O-E)^2/E

Test statistic χ2 = 9.732

c) P-value:

Degree of freedom, df = k-1

df = 5 - 1 = 4

P-Value is .0452

Since, P-Value is 0.0452.

The first option is correct.

0.025 < P-value (0.0452) < 0.05

d) Conclusion:

Since, P-value (0.0452) < α (0.10)

Reject the null hypothesis Ho. There is sufficient evidence to conclude that the not all population proportions are equal to 0.20.


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