1. Lesson 14 - 1 Test for Goodness of Fit One-Way Tables

2. Knowledge Objectives Describe the situation for which the chi-square test for goodness of fit is appropriate Define the χ 2 statistic, and identify the number of degrees of freedom it is based on, for the χ 2 goodness of fit test List the conditions that need to be satisfied in order to conduct a test χ 2 for goodness of fit Identify three main properties of the chi-square density curve

3. Construction Objectives Conduct a χ 2 test for goodness of fit. Use technology to conduct a χ 2 test for goodness of fit. If a χ 2 statistic turns out to be significant, discuss how to determine which observations contribute the most to the total value.

4. Vocabulary Goodness-of-fit test – an inferential procedure used to determine whether a categorical frequency distribution follows a claimed distribution. Expected counts – probability of an outcome times the sample size for k mutually exclusive outcomes One-way table – a table of k mutually exclusive observed values Cells – one item in the one-way table

5. Chi-Square Distributions

6. Chi-Square Distribution Total area under a chi-square curve is equal to 1 It is not symmetric, it is skewed right The shape of the chi-square distribution depends on the degrees of freedom (just like t-distribution) As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric The values of χ² are nonnegative; that is, values of χ² are always greater than or equal to zero (0); they increase to a peak and then asymptotically approach 0 Table D in the back of the book gives critical values

7. Chi-Square Test for Goodness of Fit

8. Conditions Goodness-of-fit test: Independent SRSs All expected counts are greater than or equal to 1 (all E i ≥ 1) No more than 20% of expected counts are less than 5 Remember it is the expected counts, not the observed that are critical conditions

9. Critical Region Reject null hypothesis, if P-value < α χ 2 0 > χ 2 α, k-1 P-Value is the area highlighted χ2αχ2α P-value = P(χ 2 0 ) Goodness-of-Fit Test where O i is observed count for ith category and E i is the expected count for the ith category ( O i – E i ) 2 Test Statistic: χ 2 0 = ------------- E i Σ (Right-Tailed)

10. Things to Avoid

11. Example 1 Are you more likely to have a motor vehicle collision when using a cell phone? A study of 699 drivers who were using a cell-phone when they were involved in a collision examined this question. These drivers made 26,798 cell phone calls during a 14 month study period. Each of the 699 collisions was classified in various ways. Are accidents equally likely to occur on any day of the week? SunMonTueWedThuFriSat 2013312615913611312

12. Example 1 – Graphical Analysis Are accidents equally likely to occur on any day of the week? SunMonTueWedThuFriSat 2013312615913611312

13. Example 1 – Chi-Square Analysis Are accidents equally likely to occur on any day of the week? Hypotheses: Conditions: H 0 : Motor vehicle accidents involving cell phones are equally likely to occur everyday of the week H a : Motor vehicle accidents involving cell phones will vary everyday of the week (not all the same) Expected counts (everyday) = 699/7 = 99.857 1)All expected counts > 0 2)All expected counts > 5

14. Example 1 – Chi-Square Analysis Are accidents equally likely to occur on any day of the week? Calculations: Interpretation: ItemSunMonTueWedThuFriSat Observed2013312615913611312 Expected99.86 χ²63.8611.006.8435.0313.081.7377.30 (Obs – Exp)² χ² = ∑ ----------------- Exp χ² = ∑ (63.86 + 11 + 6.84 + 35.03 + 13.08 + 1.73 + 77.3) = 208.84 Since our χ² value is much greater than the critical value (208 > 24.1), we would reject H0 and conclude that the accidents are not equally likely each day of the week. χ² (n-1,p-value) = χ² (6, 0.0005) = 24.1

15. Example 2 YellowOrangeRedGreenBrownBlueTotals Sample 1668838595396400 Sample 21094169755 Peanut0.150.230.120.150.120.231 Plain0.140.20.130.160.130.241 K = 6 classes (different colors) CS(5,.1)CS(5,.05)CS(5,.025)CS(5,.01) 9.23611.07112.83315.086

16. Example 2 (sample 1) H 0 : H 1 : Test Statistic Critical Value: Conclusion: YellowOrangeRedGreenBrownBlueTotals Observed668838595396400 Expected609248604892400 Chi-value0.60.1742.6320.0170.5210.1744.118 ( O i – E i ) 2 Test Statistic: χ 2 0 = ------------- E i Σ The big bag came from Peanut M&Ms The big bag did not come from Peanut M&Ms All critical values are bigger than 9! FTR H 0, not sufficient evidence to conclude bag is not peanut M&M’s

17. Example 2 (sample 2) H 0 : H 1 : Test Statistic Critical Value: Conclusion: YellowOrangeRedGreenBrownBlueTotals Observed1094169755 Expected8.2512.656.68.256.612.65400 Chi-value0.3711.0531.0247.2800.8732.52413.125 ( O i – E i ) 2 Test Statistic: χ 2 0 = ------------- E i Σ The snack bag came from Peanut M&Ms The snack bag did not come from Peanut M&Ms All critical values are less than 13, except for α = 0.01 ! Rej H 0, sufficient evidence to conclude bag is not peanut M&M’s

18. TI & Chi-Square (One-Way) Enter Observed values in L1 Enter Expected values in L2 Enter L3 by L3 = (L1 – L2)^2/L2 Use sum function under the LIST menu to find the sum of L3. This is the value of the χ² test statistic Largest values in L3 are the observations that are the largest contributors to the total value

19. Summary and Homework Summary –Goodness-of-fit tests apply to situations where there are a series of independent trials, and each trial has 3 or more possible outcomes –The test statistic to be used combines all of the outcomes and all of the expected counts –The test statistic has approximately a chi-square distribution –Calculator is a tool for one-way tables not a crutch! Homework –pg 846 14.1 – 14.6, 14.8