chi-square goodness-of-fit test.

Tutor: None Selected Time limit: 1 Day

Describe the chi-square goodness-of-fit test.

Provide a detailed explanation of what this test measures, and how it is similar to and different from the independent t-test and the chi-square test of independence.

How do you know when to use one analysis over the other?  Provide a real-world example.

Jun 25th, 2015

Thank you for the opportunity to help you with your question!

Chi-Square Goodness of Fit Test

When an analyst attempts to fit a statistical model to observed data, he or she may wonder how well the model actually reflects the data. How "close" are the observed values to those which would be expected under the fitted model? One statistical test that addresses this issue is the chi-square goodness of fit test. This test is commonly used to test association of variables in two-way tables (see "Two-Way Tables and the Chi-Square Test"), where the assumed model of independence is evaluated against the observed data. In general, the chi-square test statistic is of the form


A new casino game involves rolling 3 dice. The winnings are directly proportional to the total number of sixes rolled. Suppose a gambler plays the game 100 times, with the following observed counts:
Number of Sixes  Number of Rolls 
  0  48
  1  35
  2  15
  3  3
The casino becomes suspicious of the gambler and wishes to determine whether the dice are fair. What do they conclude?

If a die is fair, we would expect the probability of rolling a 6 on any given toss to be 1/6. Assuming the 3 dice are independent (the roll of one die should not affect the roll of the others), we might assume that the number of sixes in three rolls is distributed Binomial(3,1/6). To determine whether the gambler's dice are fair, we may compare his results with the results expected under this distribution. The expected values for 0, 1, 2, and 3 sixes under the Binomial(3,1/6) distribution are the following:

Null Hypothesis: 
p1 = P(roll 0 sixes) = P(X=0) = 0.58 
p2 = P(roll 1 six) = P(X=1) = 0.345 
p3 = P(roll 2 sixes) = P(X=2) = 0.07 
p4 = P(roll 3 sixes) = P(X=3) = 0.005.

Since the gambler plays 100 times, the expected counts are the following:

Number of Sixes  Expected Counts  Observed Counts 
  0  58  48
  1  34.5  35
  2  7  15
  3  0.5  3
.best of luck...............................................

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 25th, 2015

Jun 25th, 2015
Jun 25th, 2015
Oct 21st, 2016
Mark as Final Answer
Unmark as Final Answer
Final Answer

Secure Information

Content will be erased after question is completed.

Final Answer