Monday 26 November 2012

CHI SQUARE

 CHI SQUARE

Types of Data:

There are basically two types of random variables and they yield two types of data: numerical and categorical. A chi square (X2) statistic is used to investigate whether distributions of categorical variables differ from one another. Basically categorical variable yield data in the categories and numerical variables yield data in numerical form. Responses to such questions as "What is your major?" or Do you own a car?" are categorical because they yield data such as "biology" or "no." In contrast, responses to such questions as "How tall are you?" or "What is your G.P.A.?" are numerical. Numerical data can be either discrete or continuous. The table below may help you see the differences between these two variables.
 Data Type  Question Type Possible Responses
 Categorical  What is your sex? male or female
 Numerical Disrete- How many cars do you own? two or three
 Numerical Continuous - How tall are you?  72 inches
Notice that discrete data arise fom a counting process, while continuous data arise from a measuring process.
The Chi Square statistic compares the tallies or counts of categorical responses between two (or more) independent groups. (note: Chi square tests can only be used on actual numbers and not on percentages, proportions, means, etc.)

2 x 2 Contingency Table

There are several types of chi square tests depending on the way the data was collected and the hypothesis being tested. We'll begin with the simplest case: a 2 x 2 contingency table. If we set the 2 x 2 table to the general notation shown below in Table 1, using the letters a, b, c, and d to denote the contents of the cells, then we would have the following table:
Table 1. General notation for a 2 x 2 contingency table.
Variable 1
 Variable 2
 Data type 1
 Data type 2
 Totals
 Category 1
 a
b
a + b
 Category 2
 c
d
c + d
 Total
a + c
b + d
a + b + c + d = N
For a 2 x 2 contingency table the Chi Square statistic is calculated by the formula:
Note: notice that the four components of the denominator are the four totals from the table columns and rows.
Suppose you conducted a drug trial on a group of animals and you hypothesized that the animals receiving the drug would show increased heart rates compared to those that did not receive the drug. You conduct the study and collect the following data:
Ho: The proportion of animals whose heart rate increased is independent of drug treatment.
Ha: The proportion of animals whose heart rate increased is associated with drug treatment.

Table 2. Hypothetical drug trial results.
   Heart Rate
 Increased
 No Heart Rate
 Increase
Total
 Treated  36  14  50
 Not treated  30  25  55
 Total  66  39  105
Applying the formula above we get:
Chi square = 105[(36)(25) - (14)(30)]2 / (50)(55)(39)(66) = 3.418
Before we can proceed we eed to know how many degrees of freedom we have. When a comparison is made between one sample and another, a simple rule is that the degrees of freedom equal (number of columns minus one) x (number of rows minus one) not counting the totals for rows or columns. For our data this gives (2-1) x (2-1) = 1.
We now have our chi square statistic (x2 = 3.418), our predetermined alpha level of significance (0.05), and our degrees of freedom (df = 1). Entering the Chi square distribution table with 1 degree of freedom and reading along the row we find our value of x2 (3.418) lies between 2.706 and 3.841. The corresponding probability is between the 0.10 and 0.05 probability levels. That means that the p-value is above 0.05 (it is actually 0.065). Since a p-value of 0.65 is greater than the conventionally accepted significance level of 0.05 (i.e. p > 0.05) we fail to reject the null hypothesis. In other words, there is no statistically significant difference in the proportion of animals whose heart rate increased.
What would happen if the number of control animals whose heart rate increased dropped to 29 instead of 30 and, consequently, the number of controls whose hear rate did not increase changed from 25 to 26? Try it. Notice that the new x2 value is 4.125 and this value exceeds the table value of 3.841 (at 1 degree of freedom and an alpha level of 0.05). This means that p < 0.05 (it is now0.04) and we reject the null hypothesis in favor of the alternative hypothesis - the heart rate of animals is different between the treatment groups. When p < 0.05 we generally refer to this as a significant difference.
Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. For example, if, according to Mendel's laws, you expected 10 of 20 offspring from a cross to be male and the actual observed number was 8 males, then you might want to know about the "goodness to fit" between the observed and expected. Were the deviations (differences between observed and expected) the result of chance, or were they due to other factors. How much deviation can occur before you, the investigator, must conclude that something other than chance is at work, causing the observed to differ from the expected. The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result.

The formula for calculating chi-square ( 2) is:
2= (o-e)2/e
That is, chi-square is the sum of the squared difference between observed (o) and the expected (e) data (or the deviation, d), divided by the expected data in all possible categories.

For example, suppose that a cross between two pea plants yields a population of 880 plants, 639 with green seeds and 241 with yellow seeds. You are asked to propose the genotypes of the parents. Your hypothesis is that the allele for green is dominant to the allele for yellow and that the parent plants were both heterozygous for this trait. If your hypothesis is true, then the predicted ratio of offspring from this cross would be 3:1 (based on Mendel's laws) as predicted from the results of the Punnett square (Figure B. 1).


Figure B.1 - Punnett Square. Predicted offspring from cross between green and yellow-seeded plants. Green (G) is dominant (3/4 green; 1/4 yellow).









To calculate 2 , first determine the number expected in each category. If the ratio is 3:1 and the total number of observed individuals is 880, then the expected numerical values should be 660 green and 220 yellow.


Chi-square requires that you use numerical values, not percentages or ratios.


Then calculate 2 using this formula, as shown in Table B.1. Note that we get a value of 2.668 for 2. But what does this number mean? Here's how to interpret the 2 value:

1. Determine degrees of freedom (df). Degrees of freedom can be calculated as the number of categories in the problem minus 1. In our example, there are two categories (green and yellow); therefore, there is I degree of freedom.
2. Determine a relative standard to serve as the basis for accepting or rejecting the hypothesis. The relative standard commonly used in biological research is p > 0.05. The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting). In this case, using p > 0.05, you would expect any deviation to be due to chance alone 5% of the time or less.
3. Refer to a chi-square distribution table (Table B.2). Using the appropriate degrees of 'freedom, locate the value closest to your calculated chi-square in the table. Determine the closestp (probability) value associated with your chi-square and degrees of freedom. In this case (2=2.668), the p value is about 0.10, which means that there is a 10% probability that any deviation from expected results is due to chance only. Based on our standard p > 0.05, this is within the range of acceptable deviation. In terms of your hypothesis for this example, the observed chi-squareis not significantly different from expected. The observed numbers are consistent with those expected under Mendel's law.
Step-by-Step Procedure for Testing Your Hypothesis and Calculating Chi-Square
1. State the hypothesis being tested and the predicted results. Gather the data by conducting the proper experiment (or, if working genetics problems, use the data provided in the problem).
2. Determine the expected numbers for each observational class. Remember to use numbers, not percentages.


Chi-square should not be calculated if the expected value in any category is less than 5.

3. Calculate 2 using the formula. Complete all calculations to three significant digits. Round off your answer to two significant digits.
4. Use the chi-square distribution table to determine significance of the value.
  1. Determine degrees of freedom and locate the value in the appropriate column.
  2. Locate the value closest to your calculated 2 on that degrees of freedom df row.
  3. Move up the column to determine the p value.
5. State your conclusion in terms of your hypothesis.
  1. If the p value for the calculated 2 is p > 0.05, accept your hypothesis. 'The deviation is small enough that chance alone accounts for it. A p value of 0.6, for example, means that there is a 60% probability that any deviation from expected is due to chance only. This is within the range of acceptable deviation.
  2. If the p value for the calculated 2 is p < 0.05, reject your hypothesis, and conclude that some factor other than chance is operating for the deviation to be so great. For example, a p value of 0.01 means that there is only a 1% chance that this deviation is due to chance alone. Therefore, other factors must be involved.
The chi-square test will be used to test for the "goodness to fit" between observed and expected data from several laboratory investigations in this lab manual.

Table B.1
Calculating Chi-Square

  Green Yellow
Observed (o) 639 241
Expected (e) 660 220
Deviation (o - e) -21 21
Deviation2 (d2) 441 441
d2/e 0.668 2
2 = d2/e = 2.668 . .


Table B.2
Chi-Square Distribution

Degrees of
Freedom
(df)

Probability (p)
  0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001
1
0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83
2
0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
3
0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
4
0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 18.47
5
1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
6
1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46
7
2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32
8
2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12
9
3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88
10
3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59
 
Nonsignificant
Significant

No comments: