![]() According to the standard chi-square distribution table, we see that the critical value for chi-square is 7.815. Using an alpha level of 0.05, we look under the column for 0.05 and the row for degrees of freedom, which, again, is 3. Ha:O≠E (There is a statistically significant difference between observed and expected frequencies.)Īlso, the number of degrees of freedom for this test is 3. H 0:O=E (There is no statistically significant difference between observed and expected frequencies.) Therefore, our alternative hypothesis would state that there is a significant difference between the observed and expected frequencies. ![]() Since our research question asks, “Do 11th grade students prefer a certain type of lunch?” our null hypothesis for the chi-square test would state that there is no difference between the observed and the expected frequencies. First, we will need to state the null and alternative hypotheses for our research question. Let’s use our original example to create and test a hypothesis using the goodness-of-fit chi-square test. For example, if a certain card or number on a die shows up more than expected (a high observed frequency compared to the expected frequency), officials use the goodness-of-fit test to determine the likelihood that the player may be cheating or that the game may not be fair. Interestingly, goodness-of-fit tests are also used in casinos to determine if there is cheating in games of chance, such as cards or dice. There are many situations that use the goodness-of-fit test, including surveys, taste tests, and analysis of behaviors. Using our example about the preferences for types of school lunches, we calculate the degrees of freedom as follows: The number of degrees of freedom associated with a particular chi-square test is equal to the number of categories minus one. The test requires that the data are obtained through a random sample. Since there are 100 observations and 4 categories, the expected frequency of each category is 1004, or 25.Īs mentioned, the goodness-of-fit test is used to determine patterns of distinct categorical variables. To calculate the expected frequency of each category when assuming school lunch preferences are distributed equally, we divide the number of observations by the number of categories. If there is no difference in which type of lunch is preferred, we would expect the students to prefer each type of lunch equally. Using a sample of 100 11th grade students, we recorded the following information: Frequency of Type of School Lunch Chosen by Students Type of Lunch Research Question: Do 11th grade students prefer a certain type of lunch? We could construct the following table, known as a contingency table, to compare the observed and expected values. ![]() For this type of comparison, it helps to make a table to visualize the problem. We would use the chi-square goodness-of-fit test to evaluate if there was a preference in the type of lunch that 11th grade students bought in the cafeteria.
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