YOU WERE LOOKING FOR: The Chi Square Test Lab Answers
This richness of detail allows the researcher to understand the results and thus to derive more detailed information from this statistic than from many others. The Chi-square is a significance statistic, and should be followed with a strength...
The Chi-square test is a non-parametric statistic, also called a distribution free test. Non-parametric tests should be used when any one of the following conditions pertains to the data: The level of measurement of all the variables is nominal or...
If, for example, the same subjects are tested over time such that the comparisons are of the same subjects at Time 1, Time 2, Time 3, etc. The study groups must be independent. This means that a different test must be used if the two groups are related. There are 2 variables, and both are measured as categories, usually at the nominal level. However, data may be ordinal data. Interval or ratio data that have been collapsed into ordinal categories may also be used. While Chi-square has no rule about limiting the number of cells by limiting the number of categories for each variable , a very large number of cells over 20 can make it difficult to meet assumption 6 below, and to interpret the meaning of the results.
This assumption is most likely to be met if the sample size equals at least the number of cells multiplied by 5. This requirement will be fully explained in the example of the calculation of the statistic in the case study example. Many employees have contracted pneumonia leading to productivity problems due to sick leave from the disease. There is a vaccine for pneumococcal pneumonia, and the owner believes that it is important to get as many employees vaccinated as possible. Due to a production problem at the company that produces the vaccine, there is only enough vaccine for half the employees. In effect, there are two groups; employees who received the vaccine and employees who did not receive the vaccine.
The company sent a nurse to every employee who contracted pneumonia to provide home health care and to take a sputum sample for culture to determine the causative agent. They kept track of the number of employees who contracted pneumonia and which type of pneumonia each had. The dependent variable is health outcome with three levels: contracted pneumoccal pneumonia; contracted another type of pneumonia; and did not contract pneumonia. The company wanted to know if providing the vaccine made a difference.
To answer this question, they must choose a statistic that can test for differences when all the variables are nominal. Table 1 Results of the vaccination program. Health Outcome.
These phenotypes and numbers are entered in Columns 1 and 2 of the following Table 2. Your Tentative Hypothesis: This ear of corn was produced by a dihybrid cross PpSs x PpSs involving two pairs of heterozygous genes resulting in a theoretical expected ratio of See dihybrid cross in Table 1. Objective: Test your hypothesis using chi square and probability values. In order to test your hypothesis you must fill in the columns in the following Table 2.
For the observed number Column 2 , enter the number of each grain phenotype counted on the ear of corn. To calculate the observed ratio Column 3 , divide the number of each grain phenotype by 21 the grain phenotype with the lowest number of grains. For the expected ratio Column 4 , use , the theoretical ratio for a dihybrid cross. To calculate the expected number Column 5 , multiply the number of each grain phenotype by the expected fractional ratio for that grain phenotype. In the last column Column 6 , for each grain phenotype take the observed number of grains Column 2 and subtract the expected number Column 5 , square this difference, and then divide by the expected number Column 5.
Round off to three decimal places. To calculate the chi square value, add up the four decimal values in the last column Column 6.
Boston University School of Public Health Introduction This module will continue the discussion of hypothesis testing, where a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The specific tests considered here are called chi-square tests and are appropriate when the outcome is discrete dichotomous, ordinal or categorical. For example, in some clinical trials the outcome is a classification such as hypertensive, pre-hypertensive or normotensive.
We could use the same classification in an observational study such as the Framingham Heart Study to compare men and women in terms of their blood pressure status - again using the classification of hypertensive, pre-hypertensive or normotensive status. The technique to analyze a discrete outcome uses what is called a chi-square test. Specifically, the test statistic follows a chi-square probability distribution. We will consider chi-square tests here with one, two and more than two independent comparison groups.
Learning Objectives After completing this module, the student will be able to: Perform chi-square tests by hand Appropriately interpret results of chi-square tests Identify the appropriate hypothesis testing procedure based on type of outcome variable and number of samples Tests with One Sample, Discrete Outcome Here we consider hypothesis testing with a discrete outcome variable in a single population. Discrete variables are variables that take on more than two distinct responses or categories and the responses can be ordered or unordered i. The procedure we describe here can be used for dichotomous exactly 2 response options , ordinal or categorical discrete outcomes and the objective is to compare the distribution of responses, or the proportions of participants in each response category, to a known distribution.
The known distribution is derived from another study or report and it is again important in setting up the hypotheses that the comparator distribution specified in the null hypothesis is a fair comparison. The comparator is sometimes called an external or a historical control. In one sample tests for a discrete outcome, we set up our hypotheses against an appropriate comparator.
We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size n and the proportions of participants in each response category , , We then determine the appropriate test statistic for the hypothesis test. The formula for the test statistic is given below. The observed frequencies are those observed in the sample and the expected frequencies are computed as described below. The test above statistic formula above is appropriate for large samples, defined as expected frequencies of at least 5 in each of the response categories.
These expected frequencies are determined by allocating the sample to the response categories according to the distribution specified in H0. This is done by multiplying the observed sample size n by the proportions specified in the null hypothesis p 10 , p 20 , To ensure that the sample size is appropriate for the use of the test statistic above, we need to ensure that the following: min np10 , n p20 , As the name indicates, the idea is to assess whether the pattern or distribution of responses in the sample "fits" a specified population external or historical distribution.
In the next example we illustrate the test. As we work through the example, we provide additional details related to the use of this new test statistic. Example: A University conducted a survey of its recent graduates to collect demographic and health information for future planning purposes as well as to assess students' satisfaction with their undergraduate experiences. The survey revealed that a substantial proportion of students were not engaging in regular exercise, many felt their nutrition was poor and a substantial number were smoking. The next year the University launched a health promotion campaign on campus in an attempt to increase health behaviors among undergraduates. The program included modules on exercise, nutrition and smoking cessation.
To evaluate the impact of the program, the University again surveyed graduates and asked the same questions. The survey was completed by graduates and the following data were collected on the exercise question:.
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