4 One-Sample Chi-Square
4.1 The Worked Example
Here is our data:
| Values | A | B | C | D |
|---|---|---|---|---|
| Observed | 4 | 5 | 8 | 15 |
And if we add on a column showing the total number of participants, adding all the numbers in the different conditions together, (i.e. 4 + 5 + 8 + 15 = 32), then we get:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 4 | 5 | 8 | 15 | 32 |
Now the formula for the chi-square is:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
The Expected values for each condition, in a one-sample chi-square assuming a uniform (equal) distribution is calculated by \(N \times \frac{1}{k}\) where \(k\) is the number of conditions and \(N\) is the total number of participants. This can also be written more straightforward as \(N/k\). That means that in our example the expected value in each condition would be:
\[Expected = \frac{N}{k} = \frac{32}{4} = 8\]
Let's now add those Expected values to our table which looks like:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 4 | 5 | 8 | 15 | 32 |
| Expected | 8 | 8 | 8 | 8 | 32 |
We now have our data, let's start putting it into the formula, which we said was:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
Which really means:
\[\chi^2 = \frac{(Observed_{A} - Expected_{A})^2}{Expected_{A}} + \frac{(Observed_{B} - Expected_{B})^2}{Expected_{B}} + \frac{(Observed_{C} - Expected_{C})^2}{Expected_{C}} + \\ \frac{(Observed_{D} - Expected_{D})^2}{Expected_{D}}\]
So
\[\chi^2 = \frac{(4 - 8)^2}{8}+\frac{(5 - 8)^2}{8}+\frac{(8 - 8)^2}{8}+\frac{(15 - 8)^2}{8}\]
Which becomes:
\[\chi^2 = \frac{(-4)^2}{8} + \frac{(-3)^2}{8} + \frac{(0)^2}{8} + \frac{(7)^2}{8}\]
And if we now square the top halves (the numerators):
\[\chi^2 = \frac{16}{8} + \frac{9}{8} + \frac{0}{8} + \frac{49}{8}\]
Then divide the top half by the bottom half for each condition:
\[\chi^2 = {2}+{1.125}+{0}+{6.125}\] And finally add them altogether
\[\chi^2 = 9.25\]
So we find that \(\chi^2 = 9.25\)
The degrees of freedom in this test is \(k - 1\) and given that we have 4 conditions:
\[df = k - 1\] \[df = 4 - 1\] \[df = 3\]
The effect size
A common effect size for the one-sample chi-square test is \(\phi\) (pronounced "ph-aye" and can be written as "phi"). The formula for \(\phi\) is:
\[\phi = \sqrt\frac{\chi^2}{N}\]
And if we know that \(\chi^2 =9.25\) and that \(N = 32\), then putting them into the formula we get:
\[\phi = \sqrt\frac{9.25}{32}\]
\[\phi = 0.5376453\]
The write-up
If we were to look at a critical value look-up table, we would see that the critical value associated with a \(df = 3\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 7.815\). As the chi-square value of this test (i.e. \(\chi^2 = 9.25\)) is larger than \(\chi^2_{crit}\) then we can say that our test is significant, and as such would be written up as \(\chi^2(df = 3, N = 32) = 9.25,p < .05\).
Finally, if our test was significant then all we need to do is state the condition with the highest frequency (i.e. the mode), which in this case is Condition D
4.2 Test Yourself
4.2.1 DataSet 1
Here is our data:
| Values | A | B | C | D |
|---|---|---|---|---|
| Observed | 6 | 0 | 1 | 11 |
And if we add on a column showing the total number of participants, adding all the numbers in the different conditions together, (i.e. 6 + 0 + 1 + 11 = 18), then we get:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 6 | 0 | 1 | 11 | 18 |
Now the formula for the chi-square is:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
The Expected values for each condition, in a one-sample chi-square assuming a uniform (equal) distribution is calculated by \(N \times \frac{1}{k}\) where \(k\) is the number of conditions and \(N\) is the total number of participants. This can also be written more straightforward as \(N/k\). That means that in our example the expected value in each condition would be:
\[Expected = \frac{N}{k} = \frac{18}{4} = 4.5\]
Let's now add those Expected values to our table which looks like:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 6.0 | 0.0 | 1.0 | 11.0 | 18 |
| Expected | 4.5 | 4.5 | 4.5 | 4.5 | 18 |
We now have our data, let's start putting it into the formula, which we said was:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
So
\[\chi^2 = \frac{(6 - 4.5)^2}{4.5}+\frac{(0 - 4.5)^2}{4.5}+\frac{(1 - 4.5)^2}{4.5}+\frac{(11 - 4.5)^2}{4.5}\]
Which becomes:
\[\chi^2 = \frac{(1.5)^2}{4.5} + \frac{(-4.5)^2}{4.5} + \frac{(-3.5)^2}{4.5} + \frac{(6.5)^2}{4.5}\]
And if we now square the top halves (the numerators):
\[\chi^2 = \frac{2.25}{4.5} + \frac{20.25}{4.5} + \frac{12.25}{4.5} + \frac{42.25}{4.5}\]
Then divide the top half by the bottom half for each condition:
\[\chi^2 = {0.5}+{4.5}+{2.7222222}+{9.3888889}\] And finally add them altogether
\[\chi^2 = 17.1111111\]
So we find that \(\chi^2 = 17.1111111\)
The degrees of freedom in this test is \(k - 1\) and given that we have 4 conditions:
\[df = k - 1\] \[df = 4 - 1\] \[df = 3\]
The effect size
A common effect size for the one-sample chi-square test is \(\phi\) (pronounced "ph-aye" and can be written as "phi"). The formula for \(\phi\) is:
\[\phi = \sqrt\frac{\chi^2}{N}\]
And if we know that \(\chi^2 =17.1111111\) and that \(N = 18\), then putting them into the formula we get:
\[\phi = \sqrt\frac{17.1111111}{18}\]
\[\phi = 0.974996\]
The write-up
If we were to look at a critical value look-up table, we would see that the critical value associated with a \(df = 3\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 7.815\). As the chi-square value of this test (i.e. \(\chi^2 = 17.1111111\)) is larger than \(\chi^2_{crit}\) then we can say that our test is significant, and as such would be written up as \(\chi^2(df = 3, N = 18) = 17.1111111,p < .05\).
Finally, if our test was significant then all we need to do is state the condition with the highest frequency (i.e. the mode), which in this case is Condition D
4.2.2 DataSet 2
Here is our data:
| Values | A | B | C | D |
|---|---|---|---|---|
| Observed | 12 | 16 | 11 | 14 |
And if we add on a column showing the total number of participants, adding all the numbers in the different conditions together, (i.e. 12 + 16 + 11 + 14 = 53), then we get:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 12 | 16 | 11 | 14 | 53 |
Now the formula for the chi-square is:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
The Expected values for each condition, in a one-sample chi-square assuming a uniform (equal) distribution is calculated by \(N \times \frac{1}{k}\) where \(k\) is the number of conditions and \(N\) is the total number of participants. This can also be written more straightforward as \(N/k\). That means that in our example the expected value in each condition would be:
\[Expected = \frac{N}{k} = \frac{53}{4} = 13.25\]
Let's now add those Expected values to our table which looks like:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 12.00 | 16.00 | 11.00 | 14.00 | 53 |
| Expected | 13.25 | 13.25 | 13.25 | 13.25 | 53 |
We now have our data, let's start putting it into the formula, which we said was:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
So
\[\chi^2 = \frac{(12 - 13.25)^2}{13.25}+\frac{(16 - 13.25)^2}{13.25}+\frac{(11 - 13.25)^2}{13.25}+\frac{(14 - 13.25)^2}{13.25}\]
Which becomes:
\[\chi^2 = \frac{(-1.25)^2}{13.25} + \frac{(2.75)^2}{13.25} + \frac{(-2.25)^2}{13.25} + \frac{(0.75)^2}{13.25}\]
And if we now square the top halves (the numerators):
\[\chi^2 = \frac{1.5625}{13.25} + \frac{7.5625}{13.25} + \frac{5.0625}{13.25} + \frac{0.5625}{13.25}\]
Then divide the top half by the bottom half for each condition:
\[\chi^2 = {0.1179245}+{0.5707547}+{0.3820755}+{0.0424528}\] And finally add them altogether
\[\chi^2 = 1.1132075\]
So we find that \(\chi^2 = 1.1132075\)
The degrees of freedom in this test is \(k - 1\) and given that we have 4 conditions:
\[df = k - 1\] \[df = 4 - 1\] \[df = 3\]
The effect size
A common effect size for the one-sample chi-square test is \(\phi\) (pronounced "ph-aye" and can be written as "phi"). The formula for \(\phi\) is:
\[\phi = \sqrt\frac{\chi^2}{N}\]
And if we know that \(\chi^2 =1.1132075\) and that \(N = 53\), then putting them into the formula we get:
\[\phi = \sqrt\frac{1.1132075}{53}\]
\[\phi = 0.1449273\]
The write-up
If we were to look at a critical value look-up table, we would see that the critical value associated with a \(df = 3\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 7.815\). As the chi-square value of this test (i.e. \(\chi^2 = 1.1132075\)) is smaller than \(\chi^2_{crit}\) then we can say that our test is non-significant, and as such would be written up as \(\chi^2(df = 3, N = 53) = 1.1132075,p > .05\).
Finally, if our test was significant then all we need to do is state the condition with the highest frequency (i.e. the mode), which in this case is Condition B
4.2.3 DataSet 3
Here is our data:
| Values | A | B | C | D |
|---|---|---|---|---|
| Observed | 20 | 9 | 19 | 1 |
And if we add on a column showing the total number of participants, adding all the numbers in the different conditions together, (i.e. 20 + 9 + 19 + 1 = 49), then we get:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 20 | 9 | 19 | 1 | 49 |
Now the formula for the chi-square is:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
The Expected values for each condition, in a one-sample chi-square assuming a uniform (equal) distribution is calculated by \(N \times \frac{1}{k}\) where \(k\) is the number of conditions and \(N\) is the total number of participants. This can also be written more straightforward as \(N/k\). That means that in our example the expected value in each condition would be:
\[Expected = \frac{N}{k} = \frac{49}{4} = 12.25\]
Let's now add those Expected values to our table which looks like:
| Values | A | B | C | D | Total |
|---|---|---|---|---|---|
| Observed | 20.00 | 9.00 | 19.00 | 1.00 | 49 |
| Expected | 12.25 | 12.25 | 12.25 | 12.25 | 49 |
We now have our data, let's start putting it into the formula, which we said was:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
So
\[\chi^2 = \frac{(20 - 12.25)^2}{12.25}+\frac{(9 - 12.25)^2}{12.25}+\frac{(19 - 12.25)^2}{12.25}+\frac{(1 - 12.25)^2}{12.25}\]
Which becomes:
\[\chi^2 = \frac{(7.75)^2}{12.25} + \frac{(-3.25)^2}{12.25} + \frac{(6.75)^2}{12.25} + \frac{(-11.25)^2}{12.25}\]
And if we now square the top halves (the numerators):
\[\chi^2 = \frac{60.0625}{12.25} + \frac{10.5625}{12.25} + \frac{45.5625}{12.25} + \frac{126.5625}{12.25}\]
Then divide the top half by the bottom half for each condition:
\[\chi^2 = {4.9030612}+{0.8622449}+{3.7193878}+{10.3316327}\] And finally add them altogether
\[\chi^2 = 19.8163265\]
So we find that \(\chi^2 = 19.8163265\)
The degrees of freedom in this test is \(k - 1\) and given that we have 4 conditions:
\[df = k - 1\] \[df = 4 - 1\] \[df = 3\]
The effect size
A common effect size for the one-sample chi-square test is \(\phi\) (pronounced "ph-aye" and can be written as "phi"). The formula for \(\phi\) is:
\[\phi = \sqrt\frac{\chi^2}{N}\]
And if we know that \(\chi^2 =19.8163265\) and that \(N = 49\), then putting them into the formula we get:
\[\phi = \sqrt\frac{19.8163265}{49}\]
\[\phi = 0.6359362\]
The write-up
If we were to look at a critical value look-up table, we would see that the critical value associated with a \(df = 3\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 7.815\). As the chi-square value of this test (i.e. \(\chi^2 = 19.8163265\)) is larger than \(\chi^2_{crit}\) then we can say that our test is significant, and as such would be written up as \(\chi^2(df = 3, N = 49) = 19.8163265,p < .05\).
Finally, if our test was significant then all we need to do is state the condition with the highest frequency (i.e. the mode), which in this case is Condition A