Chi-square tests #3 - Questions and Answers

Question 1 Describe a goodness-of-fit test. What assumptions are made when using a goodness-of- fit test? Answer Step 1 The chi-square is a non-parametric test which is used for comparing the categorical variables. The chi-square goodness of fit test helps in determining how the observed distribution fits the expected distribution. It is used for comparing the observed data with the expected data. In other words, it determines whether the data of the sample is reliable with the population distribution or not. Step 2 Assumptions: -The sample drawn must be randomly selected. -The variable considered for the study must be a categorical variable. -Different levels of categorical variable must be mutually exclusive. -Each of the observations from the sample must be independent of each other. -Each level of the categorical variable must have at least 5 expected frequencies. Question 2 A chi-square homogeneity test is to be conducted to decide whether four populations are nonhomogeneous with respect to a variable that has eight possible values. What are the degrees of freedom for the χ²-statistic? Answer Step 1 A categorical variable's distribution in two or more populations (or subgroups of a population) is compared using the homogeneity test to see if they share the same distribution. Step 2 For chi-square tests based on two-way tables (both the test of independence and the test of homogeneity), the degrees of freedom are calculated using the following formula: df = (r-1)(c-1) where r is the number of rows and c is the number of columns. Question 3 "Determine the critical values for these tests of a population standard deviation. (a) A right-tailed test with 18 degrees of freedom at the α = 0.01 level of significance (b) A left-tailed test for a sample of size n= 24 at the α = 0.1 level of significance (c) A two-tailed test for a sample of size n= 26 at the α = 0.01 level of significance" (a) The critical value for this right-tailed test is (Round to three decimal places as needed.) (b) The critical value for this left-tailed test is (Round to three decimal places as needed.) (c) The critical values for this two-tailed test are (Round to three decimal places as needed. Use a comma to separate answers as needed.) Answer a) The test's right-tailed nature is a given. There are 18 degrees of freedom. The significance threshold is 0.01. Excel's "=CHISQ.INV.RT (0.01,18)" formula is used to get the chi-square test's critical value. 34.805 is the right-tailed critical value. b) The test is assumed to be left-tailed. 24 people make up the sample. n-1=24-1=23 is the number of degrees of freedom. The significance threshold is 0.10. Excel's "=CHISQ.INV (0.10,23)" formula is used to get the chi-square test's critical value. 14.848 is the left-tailed critical value. c) The two-tailed nature of the test is a given. 26 people make up the sample. N-1=26-1=25 is the number of degrees of freedom. The significance threshold is 0.01. Excel's "=CHISQ.INV (0.005,25)" formula is used to get the lower critical value for the chi- square test. Excel's "=CHISQ.INV.RT (0.005,25)" formula is used to get the chi-square test's upper critical value. The critical values with two tails are (10.520, 46.928). Question 4 A bag of candy contains 5 different types of colored candies; red, green, blue, yellow, and orange. According to the manufacturer, bags should contain an equal number of each color. Students in a statistics class decided to use a chi-square procedure to test the manufacturer's claim. They opened a bag of candy and recorded the number of candies of each color. The results are shown in the following table. Red Green Blue Yellow Orange 17 24 20 25 14 Which color contributes most to the chi-square test statistic? a. Red b. Green c. Blue d. Yellow e. Orange Answer The null and alternative hypotheses are stated below: H₀: P1= P2=P3= P4= P5=1/5. Hₐ: At least one Pᵢ is not equal to 0.20. The chi-square test is appropriate in this situation. The test statistic is given below: x² = Σ(Oᵢ - Eᵢ)² / Eᵢ ~ χ²(k-1) Where Oᵢ=Observed frequency, Eᵢ= Expected frequency, k= Number of observations The Expected frequencies and test statistics can be obtained as follows: Oᵢ Eᵢ (Oᵢ - Eᵢ)²/Eᵢ Red 17 20(= 100 × 0.20) Green 24 20(= 100 × 0.20) Blue 20 20(= 100 × 0.20) Yellow 25 20(= 100 × 0.20) Orange 14 20(= 100 × 0.20) Total 100 100 From the above table, it is clear that (Oᵢ – Eᵢ)² / Eᵢ is high for orange. Thus, orange contribute most to chi-square test statistics.