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Exhibit 6.1 Chi-squared calculations via contingency table
is approximately 7.82. Thus, we can reject the null since 106.8 > 7.82. 2 Also,
the p-value from Exhibit 6.1 is extremely small (5.35687E-23) 3 indicating a very
small probability of obtaining the
2 value of 106.8 when the null hypothesis is
true. The p-value returned by the CHITEST function is shown in cell F17, and it
is the only value that is needed to reject, or not reject, the null hypothesis. Note
that the cell formula in F18 is the calculation of the
2 given in the formula above
and is not returned by the CHITEST function. This result leads us to conclude that
the null hypothesis is likely not true, so we reject the notion that the variables are
independent. Instead, there appears to be a strong dependence given our test statistic.
Earlier, we summarized the general steps in performing a test of hypothesis. Now
we describe in detail how to perform the test of hypothesis associated with the
test. The steps of the process are:
1. Organize the frequency data related to two categorical variables in a contingency
2 Tables of χ
can be found in most statistics texts. You will also need to calculate the degrees
of freedom for the data: (number of rows–1) × (number of columns–1). In our example: (2–1) ×
(4–1) = 3.
3 Recall this is a form of what is known as “scientific notation”. E-17 means 10 raised to the –17
power, or the decimal point moved 17 decimal places to the left of the current position for 3.8749.
Positive (E+13 e.g.) powers of 10 moves the decimal to the right (13 decimal places).
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