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Chi-Square Distribution

Chi-Square Distribution. Chapter 15. Chapter Goals. When you have completed this chapter, you will be able to:. Understand the nature and role of chi-square distribution. Identify a wide variety of uses of the chi-square distribution.

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Chi-Square Distribution

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  1. Chi-Square Distribution Chapter 15

  2. Chapter Goals When you have completed this chapter, you will be able to: Understand the nature and role of chi-square distribution Identify a wide variety of uses of the chi-square distribution Conduct a test of hypothesis comparing an observed frequency distribution to an expected frequency distribution and...

  3. 10 Chapter Goals Conduct a test of hypothesis for normality using the chi-square distribution Conductahypothesistestto determine whether two attributes are independent

  4. Characteristics ofthe Chi-Square Distribution … it is positively skewed … it is non-negative … it is based on degrees of freedom …when the degrees of freedom change a new distribution is created …e.g.

  5. Characteristics ofthe Chi-Square Distribution df = 3 df = 5 df = 10 c2

  6. Goodness-of-Fit Test: Equal Expected Frequencies Let f0 and febe the observed and expected frequencies respectively H0: There is no differencebetween the observed and expected frequencies H1: There is a difference between the observed and the expected frequencies

  7. ( ) é é 2 - f f å 2 o e c = ê ê f ê ê ë ë e Goodness-of-Fit Test: Equal Expected Frequencies … the test statistic is: …the critical value is a chi-square value with (k-1) degrees of freedom, where k is the number of categories

  8. Solve Goodness-of-Fit Test: Equal Expected Frequencies The following information shows the number of employees absent by day of the week at a large a manufacturing plant. DayFrequency Monday 120 Tuesday 45 Wednesday 60 Thursday 90 Friday 130 Total 445 At the .05 level of significance, is there a difference in the absence rate by day of the week?

  9. Goodness-of-Fit Test: Equal Expected Frequencies Hypothesis Test H0: There is no difference in absence rate by day of the week… Step 1 (120+45+60+90+130)/5 = 89 H1: Absence rates by day are not all equal  = 0.05 Step 2 (5-1) = 4 Degrees of freedom Use Chi-Square test Step 3 9.488. (see Appendix I) Step 4 Reject H0 if 2 > Chi-Square

  10. Reject H0 if 2 > 9.488 Using the Table… Degrees of Freedom 5 – 1 = 4 Right-Tail Area  = 0.05

  11. Day Frequency Expected (fo – fe)2/fe Monday 120 89 10.80 Tuesday 45 89 21.75 Wednesday 60 89 9.45 Thursday 90 89 0.01 Friday 130 8918.89 Total 445445 60.90 = 1.98 Test Statistics (120-89)2/89 Step 5 2 Reject H0 if 2 > 9.488 Conclusion: Reject the null hypothesis. Absentee rates are not the same for each day of the week.

  12. Solve A U.S. Bureau of the Census indicated that… Not re-married Never married A sample of 500 adults from the Philadelphia area showed: At the .05 significance level, can we conclude that the Philadelphia area is differentfrom the U.S. as a whole?

  13. … continued 2 2.3814 Status Expected (fo – fe)2/fe * Census figures would predict: i.e. 639*500 = 319.5 ** Our sample: (310-319.5)2/319.5 = .2825

  14. … continued H0: The distribution has not changed Step 1 H1: The distribution has changed.  = 0.05 Step 2 H0 is rejected if 2>7.815, df = 3 Step 3 2 = 2.3814 Step 4 Conclusion: Reject the null hypothesis. The distribution regarding marital status in Philadelphia is different from the rest of the United States.

  15. Procedure …to determine the mean and standard deviation of the frequency distribution Goodness-of-Fit Test: Normality … the test investigates ifthe observed frequencies in a frequency distribution match the theoretical normal distribution - Compute thez-value for the lower class limit and the upper class limit for each class - Determinefe for each category - Use the chi-squaregoodness-of-fit test to determine if fo coincides with fe

  16. Goodness-of-Fit Test: Normality • A sample of 500 donations to the Arthritis Foundation is reported in the following frequency distribution • Is it reasonable to conclude that the distribution is normally distributed with a mean of $10 and a standard deviation of $2? • Use the .05 significance level

  17. … continued

  18. … continued - m - z X 6 10 = = = - 2 . 00 s 2 < - = - = P ( z 2 . 00 ) 0 . 5000 . 4772 .0228 To compute fefor the first class, first determine the z - value Now… find the probability of a z - value less than –2.00

  19. … continued

  20. … continued = = f (. 0228 )( 500 ) 11 . 40 e The expected frequency is the probability of a z-value less than –2.00 times the sample size The other expected frequencies are computed similarly

  21. … continued

  22. … continued H0: The observations follow the normal distribution Step 1 H0: The observations do NOT follow the normal distribution Step 2  = 0.05 H0 is rejected if 2>7.815, df = 6 Step 3 2 = 336.33 Step 4 Conclusion: H0: is rejected. The observations do NOT follow the normal distribution

  23. Contingency Table Analysis Acontingency tableis used to investigate whether two traits or characteristicsare related … each observation is classified according to two criteria …the usual hypothesis testing procedure is used Note: … the degrees of freedomis equal to: (number of rows -1)(number of columns -1) … the expected frequency is computed as: Expected Frequency = (row total)(column total)/grand total

  24. Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the .05 level of significance, can we conclude that gender and the location of the accident are related?

  25. … continued Location The expected frequency for the work-male intersection is computed as (90)(80)/150 =48 Similarly, you can compute the expected frequencies for the other cells

  26. … continued ( ) ( ) 2 2 - - 60 48 10 8 2 c = + + ... 48 8 = 16 . 667 H0: is rejected. Gender and Location are related! Conclusion: H0: The Gender and Location are NOT related Step 1 H0: The Gender and Location are related Step 2  = 0.05 H0 is rejected if c2 >5.991, df = 2 Step 3 (…there are (3- 1)(2-1) = 2 degrees of freedom) Step 4 Find the value of c2

  27. www.mcgrawhill.ca/college/lind for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more! Test your learning… Click on… Online Learning Centre

  28. This completes Chapter 15

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