Chi-square Goodness of Fit Test

Presentation 10.1

Another Significance Test for Proportions

• But this time we want to test multiple proportions.

• Hypothesis tests can also be performed with one proportion to obtain evidence about the truth about a population.

M&Ms Again

• The Mars Company claims the color distribution at right.

• A sample from a king size bag found a distribution that did not exactly follow the one at right.

• Is our sample bag enough evidence to dispute the Mars Company’s claim about the distribution of colors?

• The chi-square test can determine this.

Chi-square Goodness of Fit Test Formulas

dfcdfValueP

categoriesofdfE

EO

ExpectedObservedH

ExpectedObservedH

a

,9999,

1#

:

:

22

22

0

Null Hypothesis

Alternate Hypothesis

Test Statistic(that symbol is called“Chi-squared”)

The null and alternate hypotheses are always the same with a Goodness of Fit Test.

O is the observed count for each category and E is the expected count for each category.

Instead of a normal or t distribution, we now have a chi-squared distribution

M&Ms Example

• Look at the data of the 86 candies in our king size bag

Color Brown Yellow Red Blue Orange Green

Observed in our bag

12 10 8 25 14 19

Supposed (expected)

%

13% 14% 13% 24% 20% 16%

Supposed (expected) count for our bag

(.13)86 =11.18

(.14)86 =12.04

(.13)86 =11.18

(.24)86 =20.64

(.20)86 =17.2

(.16)86 =13.76

Conditions for the Goodness of Fit Test

• None of the observed counts should be less than 1

• No more than 20% of the counts should be less than 5

• These are simple checks to make sure that the sample size is sufficient.

M&Ms Example

• Check the conditions– Since all counts are greater than 5, we are ok to

conduct the test– Our counts were 12, 10, 8, 25, 14, 19.

• Write Hypotheses (these are always the same!)– Null: Ho: Observed = Expected

• That is, what we observed should be the same as what we expected (what the Mars company advertises)

– Alternate: Ha: Observed ≠ Expected• That is, the color distribution is just too different from what is

advertised to be attributed to random chance.

M&Ms Example

• Calculations (this takes a bit of work)

Color Brown Yellow Red Blue Orange Green

Observed 12 10 8 25 14 19

Expected 11.18 12.04 11.18 20.64 17.2 13.76

(O-E)2 (12-11.18)2=0.6724

(10-12.04)2=4.1616

(8-11.18)2=10.1124

(25-20.64)2=19.0096

(14-17.2)2=10.24

(19-13.76)2=27.4576

(O-E)2/E 0.0601 0.3456 0.9045 0.9210 0.5953 1.9955

E

EO 22

822.4

9955.15953.921.9045.3456.0601.2

2

Then, add them up!

M&Ms Example

• Finish calculations• We have 6 categories so our

degrees of freedom: 6 – 1 = 5.• Then use X2cdf(4.822,999,5)

to find the p-value• With a p-value so high we fail

to reject the null.• There is not sufficient evidence

to suggest that the color distribution in our bag is different from what is advertised.

Chi-square Goodness of Fit Test

This concludes this presentation.