The following table records the observed number of births at a hospital in four consecutive quarterly periods:

\begin{array}{|c|c|c|c|c|} \hline Quarters & Jan-Mar & Apr-Jun & Jul-Sep & Oct-Dec \\ \hline \text{Number of births} & 110 & 57 & 53 & 80 \\ \hline \end{array}

It is conjectured that twice as many babies are born during the January-March quarter than are born in any of the other three quarters. At $\alpha = 0.05$, test if the data strongly contradicts the stated conjecture.

I have the null hypothesis being

$$H_0: p_1 = 0.4, p_2 = 0.2, p_3 = 0.2, p_4 = 0.2$$

$$H_a: \text{One of these equalities does not hold}$$

Peason's Test for Goodness of Fit gives

$$X^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i} \approx 8.47$$

However, my SAS output and the F table give contradicting results, so I think I am interpreting one incorrectly.

SAS Output:

enter image description here

This would indicate a rejection of the null hypothesis at $\alpha = 0.05$.

However, an F table with $df_1 = 2$, $df_2=3$ gives $9.5521$ which would indicate that we do not reject the null hypothesis since $8.47 \lt 9.5521$.

  • $\begingroup$ It just occured to me that the F table and Chi-Square table may not be the same. I interpreted $X_{3}^2$ as being an F value with df of 2 and 3 but the 2 was for Chi-Squared. A chi-squared table give 7.81, thus rejecting the null hypothesis. $\endgroup$ – Remy Dec 2 '17 at 22:21

Subject to the null hypothesis and based on $n = 300$ observations, the expected numbers of births in the four quarters are $E_i =120,\, E_2 = 60,\, E_3 = 60,\, E_4 = 60.$

Then the chi-squared goodness-of-fit (GOF) statistic is $Q = \sum_{i=1}^4 \frac{(X_i = E_i)^2}{E_i} = 8.467,$ as you have found.

Because all of the $E_i > 5,$ the GOF statistic $Q$ is approximately distributed as $\mathsf{Chisq}(df=4-1 = 3).$ The critical value for a test at the 5% level is $q = 7.815$ because 7.815 cuts 5% of the probability from the upper tail of the distribution $\mathsf{Chisq}(df=3).$ You can get the critical value from printed tables of the chi-squared distribution or using software. The value from R statistical software is shown below.

qchisq(.95, 3)
##  7.814728

Because $Q = 8.467 > q = 7.815$ you can reject (at level 5%) the null hypothesis that the quarterly birth distribution is $p = (.4, .2, .2, .2).$

The SAS output shows the P-value, which is the probability that a random variable with distribution $\mathsf{Chisq}(df=3)$ exceeds 8.467. The P-value 0.0373 is found using R as follows:

1 - pchisq(8.467, 3)
## 0.03728462

You are correct that F and chi-squared are two different families of distributions, and that you should not use the F-distribution for this test.

[F-distributions can be used to compare two variances or to do the main test for most analysis of variance (ANOVA) designs.]

In the figure below, the critical value $q = 7.815$ is shown as an orange dashed line; 5% of the area under the density curve is to the right of this line. The observed value of the GOF statistic $Q = 8.467$ is shown as a solid black line; the P-value is the area under the density curve to the right of this line.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.