ABSTRACT
Total 150 The null hypothesis would be that the incoming Pharm.D. students are distributed the same as previous classes based on the type on institution where they received their pre-pharmacy training. In selecting the appropriate critical χ2-value, the number of degrees of freedom is one less than the number of levels of the discrete variable (k). Once again the k – 1 degrees of freedom is selected to correct for bias. In this example, since there are four types of institutions being tested, the number of degrees of freedom is three. The decision rule, assuming 95% confidence is: with α = 0.05, reject H0 if χ2 > χ23(0.05) = 7.8147. The value 7.8147 is found in Table B15 at the intercept of the third row (degrees of freedom equal to three) and the second column of critical values (α = 0.05). If there were no differences between the observed institutions and the expected enrollment, we would expect to see similar proportions. The χ2-statistic would be calculated as follows:
Observed Expected O – E (O – E)2/E Parent institution 85 90
−5.0 0.278 In-state system school 35 37.5
−2.5 0.167 In-state non-system school 5 7.5
−2.5 0.833 Out-of-state school 25 15 +10.0 6.667 χ2 = 7.945
Based on the results of the chi square test, with the calculated χ2 greater than 7.8147; we reject the null hypothesis and conclude that there is a significant difference between the characteristics of the new students compare to what is traditionally expected in an entry class based on institution where they received their previous education. Unfortunately the chi square test does not indicate where the significant difference(s) exist between of the observed and expected results. The simplest way to estimate the major difference(s) is to observe which level(s) of the discrete variable contribute the most to a significant chi square statistic. This would be represented by the largest (O-E)2/E. In this example it would be Out-of-State Schools which contributed 83.9% (6.667/7.945) of the total deviation from expected. Alternatively, one could perform two additional chi square tests to determine if the Out-of-State School proportion is significantly different previous experience by: 1) performing a second chi square test without the data for Out-of-State Schools; and 2) performing a third chi square on Out-of-State Schools to determine if it differs significantly form the average of the other three school sources. The second chi square would be tested against a critical chi square value with two degrees of freedom (k - 1) which is 5.9915 (Table B15). The calculation would be as follows:
Observed Expected O – E (O – E)2/E Parent institution 85 90
−5.0 0.278 In-state system school 35 37.5
−2.5 0.167 In-state non-system school 5 7.5
−2.5 0.833 χ2 = 1.278
The calculated χ2 is less than 5.9915 and we fail to reject the null hypothesis that the proportions for these three types of institutions are the same as in previous years. Therefore, our best guess is that the proportions of students from the three remaining instructions are distributed similarly to previous years and only the out-of-state institutions are disproportionately represented. The third chi square would be tested against a critical chi square value with one degree of freedom comparing out-of state schools to the composite for the three other types of institutions. In this case the critical value for one degree of freedom is 3.8415 (Table B15). The calculation would be as follows:
Observed Expected O – E (O – E)2/E In-state school 125 135
−10.0 0.741 Out-of-state school 25 15 +10.0 6.667 χ2 = 7.408
Here calculated χ2 is greater than the critical value of 3.8415, so we would reject the null hypothesis and conclude that there is a significantly greater proportion of out-ofstate school students in the entering Pharm.D. class. Chi Square for One Discrete Dependent Variable and Equal Expectations If different batches of a particular product are compared, where production was similar, we could expect to see an equal result for some measure of the final product regardless of the batch tested. For example, assume that we wish to compare four lots of a particular drug for some minor undesirable trait (e.g., a blemish on the tablet coating). We randomly sample 1000 tablets from each batch and examine the tablets for that trait. The results of the experiment are as follows:
Number of Tablets with Blemishes Batch A 12 Batch B 15 Batch C 10 Batch D 9
A simple hypothesis to evaluate this data could be as follows: H0: The samples are selected from the same population H1: The samples are from different populations Our best estimate of expected frequency is the average of the sample frequencies:
Eq. 16.3 In this particular case:
11.5 = 4
9 + 10 + 15 + 12 = f E
Therefore the χ2-statistic would be calculated as follows:
Observed Expected O – E (O – E)2/E Batch A 12 11.5 +0.5 0.02 Batch B 15 11.5 +3.5 1.07 Batch C 10 11.5 –1.5 0.20 Batch D 9 11.5 –2.5 0.54 χ2 = 1.83
Based on the results of the chi square test, with the calculated χ2 less than 7.8147, we fail to reject the null hypothesis. Therefore, our best guess is that they are from the same population; in other words, there is no difference among the four batches.
