Microsoft Office Tutorials and References

In Depth Information

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

χ

α

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

χ

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

χ

2

χ

test. The steps of the process are:

1. Organize the frequency data related to two categorical variables in a contingency

table.

2
Tables of
χ

2

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).

α

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