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Test of Independence for a $2 \times 2$ Table

There are many ``goodness-of-fit'' type tests. We will look at a simple one which tests for independence in a $2 \times 2$ table.

Consider a professional basketball player. Suppose he has two free throws. Is the result of the second free throw independent of the first? This has perplexed society for ages. Our hypotheses are:

\begin{displaymath}\mbox{$H_0:\;$\space First and Second shot are independent versus
$H_A:\;$\space First and Second shot are dependent}\;.
\end{displaymath}

Complements of Kitchens (1998, Exploring Statistics, Pacific Grove, CA: Duxbury Press), we have data on two basketball players: Larry Bird and Rick Robey. Consider the data for Larry Bird (taken during the season 1980-1981).

    Second Shot  
    Hit Miss Total
First Hit 251 34 285
shot Miss 48 5 53
  Total 299 39 338

The $\chi^2$ test is extremely simple. The above table are the observed frequencies. Simply form the expected frequencies under H0 and obtain the $\chi^2$ statistic discussed above, (1). Our rejection rule changes slightly to:

 \begin{displaymath}
\mbox{Reject $H_0$\space in favor of $H_a$\space if $\chi^2 \geq 3.84$ }\;.
\end{displaymath} (3)

How do we get the expected frequencies under H0? Under H0,

\begin{displaymath}P[\mbox{Hit First and Hit Second}] =
P[\mbox{Hit First}]P[\mbox{Hit Second}] \;.
\end{displaymath}

What good does this do? Well for one thing, I can estimate the right-side by

\begin{displaymath}\widehat{P[\mbox{Hit First}]P[\mbox{Hit Second}]} = \frac{285}{338} \frac{299}{338}
=0.6498 \;,
\end{displaymath}

Now .6498 is an estimate of the left-side provided H0 is true; otherwise, it is not. Hence an estimate of the expected frequency under H0 is

\begin{displaymath}\widehat{E}_{11} = 388 \frac{285}{338} \frac{299}{338} = 252.11
\end{displaymath}

where the subscript 11 stands for the cell in row 1 and column 1. Notice how close this is to the observed frequency. To complete this table we need the other 3 expected frequencies. But wait, these expected frequencies must add to the margin frequencies; hence, we can just subtract. Once we do we get the following table with expected frequencies in parentheses:
    Second Shot  
    Hit Miss Total
First Hit 251 34 285
    (252.11) (32.88)  
shot Miss 48 5 53
    (46.88) (6.12)  
  Total 299 39 338
The computation of $\chi^2$ is

\begin{displaymath}\chi^2 = 0.005 + 0.038 +
0.027 + 0.203 = 0.273
\end{displaymath}

Because $\chi^2 \not \geq 3.84$, we fail to reject H0. It seems that Larry Bird's (at least in the season 1980-81) first and second freethrows were independent of one another.




EXERCISES

0.0.2   Below are similar data for the basketball player Rick Robey; complements of Kitchens (1998, Exploring Statistics, Pacific Grove, CA: Duxbury Press), Obtain the $\chi^2$ test of independence for this data.
    Second Shot  
    Hit Miss Total
First Hit 54 37 91
shot Miss 49 31 80
  Total 103 68 171


next up previous contents
Next: About this document ... Up: Goodness-of-Fit Tests Previous: Chi-Squared Goodness of Fit
Stat 160
2002-04-12