## Chi-Squared Goodness of Fit Test

**Overview:**

The goodness of fit test determines if an observed distribution could match a known theoretical distribution. We compare each of the observed count to the expected count.

**Hypothesis**

A hypothesis test is used to address how to determine the observed numbers that fit the sample “null” model.

**Assumptions and Conditions**

Randomization Condition: The data being collected must be random.

Expected Cell Frequency Condition: We should expect to have at least 5 individuals or entries in each cell.

**Mechanics**

χ²: Chi-squared test statistic is when you add the sum of the squares of the deviation between the observed and expected counts divided by the expected counts.

Degrees of Freedom:

In the conclusion you would either reject or fail to reject your null hypothesis. If the p-value is higher than the alpha level (0.05) then you would fail to reject the null hypothesis and there is not enough evidence to support the question presented. If the p-value is lower than the alpha level, then you would reject the null and there is enough evidence to support it.

Stat › List › Enter the observed counts into L1, and the expected counts into L2

Stat › Tests › D: χ² GOF-Test

*n*is the number of categories*n*-1**Conclusion:**In the conclusion you would either reject or fail to reject your null hypothesis. If the p-value is higher than the alpha level (0.05) then you would fail to reject the null hypothesis and there is not enough evidence to support the question presented. If the p-value is lower than the alpha level, then you would reject the null and there is enough evidence to support it.

**TI Tip:**Stat › List › Enter the observed counts into L1, and the expected counts into L2

Stat › Tests › D: χ² GOF-Test

**Example:** a) You would add 11 + 7 + 9 + 15 + 12 + 6 = (60 / 6) =10

b) The chart has only one set of counts so it is a goodness of fit test.

c) H0: The die is fair meaning all faces have a

d The rolls are random and the expected counts are bigger than 5.

e)

f) Place the counts into L1 and use the x2 GOF test statistic which would result in having a p-value of .3471 and x2 = 5.6

g) With a p-value of .3471 we fail to reject the null hypothesis, and there is no evidence to support that the die is unfair.

b) The chart has only one set of counts so it is a goodness of fit test.

c) H0: The die is fair meaning all faces have a

*p*= .17 HA: The die is not fair.d The rolls are random and the expected counts are bigger than 5.

e)

*n*-1 = 6-1 = 5f) Place the counts into L1 and use the x2 GOF test statistic which would result in having a p-value of .3471 and x2 = 5.6

g) With a p-value of .3471 we fail to reject the null hypothesis, and there is no evidence to support that the die is unfair.