Goodness of Fit

Before the Gondola was in operation, Heavenly tracked its skiers and boarders and found the following

 Type Percent of Skiers Beginner 30% Intermediate 40% Advanced 20% Expert 10%

With the new gondola in place the ski resort wants to determine if the distribution has changed.  They tracked 2000 skiers and boarders and came up with the following

 Type Observed Count Beginner 590 Intermediate 860 Advanced 400 Expert 150

What can be concluded  (use a  =  .05)?

Solution

We first write determine the null and alternative hypotheses

H0:  The new population of skiers and boarders follows the same distribution as the old distribution of skiers.

H1:  The new population of skiers and boarders does not follow the same distribution as the old distribution of skiers.

Next we compute the expected counts by multiplying the sample size 2000 by the expected percent.

 Type Observed Count Expected Count (O - E)2 (O - E)2 / E Beginner 590 600 100 0.167 Intermediate 860 800 3600 4.5 Advanced 400 400 0 0 Expert 150 200 2500 12.5

Now add up the total of the last column to get

0.167 + 4.5 + 0 + 12.5  =  17.17

The number of degrees of freedom is

df  =  n - 1  =  4 - 1  =  3

where n is the number of rows in the table.

We use the table for the Chi Square distribution.  The critical value that corresponds to a level of significance of .05 with 3 degrees of freedom is 7.81.

Since

17.17  >  7.81

we can reject the null hypothesis and accept the alternative hypothesis and conclude that the distribution of skiers and boarders has changed.

An applet that does goodness of fit computations can be found here