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13.1 The Chi-square Goodness-of-Fit test

13.1 The Chi-square Goodness-of-Fit test.

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13.1 The Chi-square Goodness-of-Fit test

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  1. 13.1 The Chi-square Goodness-of-Fit test

  2. The chi-square goodness-of-fit test is a statistics used to compare and decide whether two or more populations, variables or characteristics are the same. It does not matter what the distributions of the populations are so long as the relative frequencies are known for each population or the population and some standard population frequencies.

  3. 1. Know the procedure used to evaluate Goodness-of-Fit chi-square tests. • Problem: The number of defects for a new thermometer is classified by the following defect types with their expected defects percent obtained from historical statistics from an older model thermometer:

  4. Step 1. Make a problems statement: (becomes the hypothesis statement, Ho ). • Is the new thermometer the same quality (based on defect profile) as the old? or Are the distribution of defects from the new thermometer the same as the distribution of defects from the old?. Is the new thermometer (based on the characteristics of defect proportions) the same as the standard (old thermometer)? • Hypothesis: • Ho (null hypothesis): Old = New or Old is same as New or Distribution of New is same as Old. • Ha: Ho is not true. (alternate hypothesis) The distributions are different.

  5. Step 2. Choose, the significance level of the test. • If you want to be 95 % certain that the test is true, then alpha = 0.05

  6. Step 3. Determine or compute the expected frequencies:

  7. Step 4: Compute • = 9.72 from computational table Step 5: Perform test chi-square test:  .02 < p < .025 Step 6: Conclusion or inference: • The old thermometer is different than the new (pertaining to the characteristics of defect distribution) • So Ho (null hypothesis) is rejected; there seems to be a difference between old and new.

  8. Example #1 • The paper “Linkage Studies of the Tomato” (Trans. Royal Canada. Inst. (1931)) reported the accompanying data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate if the frequencies are consistent with genetic laws which state the phenotypes should occur in the ratio 9:3:3:1.

  9. 1. State appropriate null and alternative hypotheses for investigating the genetic model. • Ho: The proportions reported are the same as the proportions governed by genetic laws. (OR ) • Ha: At least one of the proportions reported is different from the genetic laws. (OR at least one of these proportions is incorrect, where p1 = proportion of tall cut leaf offspring, p2 = the proportion of tall pot leaf offspring, p3 = the proportion of dwarf cut leaf offspring, and p4 = the proportion of dwarf pot leaf offspring.)

  10. 2. Verify conditions for performing inference in this setting. • We need to be able to consider these offspring as an SRS of potential offspring from the genetic crossing. If not, we may not be able to generalize our results. All expected counts are far greater than 5, so we are safe to use the chi-square goodness of fit test.

  11. 3. Construct an appropriate graphical display to compare the results of the experiment with the hypothesized values.

  12. 4. Calculate the test statistic and the P-value. 5. What conclusion would you draw?

  13. 4. Calculate the test statistic and the P-value. • P > 25% • P = 5. What conclusion would you draw? Since our p value is so large, we do not have enough evidence to reject Ho. That is, we do not have enough evidence to suggest that the reported proportions differ from the proportions given by genetic laws.

  14. Example #2 • The Advanced Placement (AP) Statistics examination was first administered in May 1997. Students’ papers are graded on a scale of 1–5, with 5 being the highest score. Over 7,600 students took the exam in the first year, and the distribution of scores was as follows (not including exams that were scored late). • Score 5 4 3 2 1 . • Percent 15.3 22.0 24.8 19.8 18.1 • A distance learning class that took AP Statistics via satellite television had the following distribution of grades: • Score 5 4 3 2 1 . • Frequency 7 13 7 6 2

  15. 1. Calculate marginal percents and make a bar graph of the population scores and the sample scores on the same grid, so that the two distributions can be compared visually.

  16. 2. Carry out an appropriate test to determine if the distribution of scores for students enrolled in the distance learning program is significantly different from the distribution of scores for all students who took the inaugural exam.

  17. We must be willing to treat this class of students as an SRS from the population of all distance learning classes. We will proceed with caution. All expected counts are 5 or more. • Ho: The distribution of AP Statistics exams scores for distance learning students is the same as the distribution of scores for all students who took the May 1997 exam. • Ha: The distribution of AP Statistics exams scores for distance learning students is different than the distribution of scores for all students who took the May 1997 exam.

  18. We will use a significance level of 0.05. There are 5 categories, meaning there are 5 – 1 = 4 degrees of freedom. • .10 < P < .15. • We do not have enough evidence to reject Ho since p > alpha. We do not have enough evidence to suggest the distributions of scores of traditional students is different than the distribution of scores of the distance learning students.

  19. Example #3 • Lacking a table of random numbers, Dan decided to use the last digits in a set of five-place tables of logarithms. The number of times each integer occurred in a sample of size n = 50 is given below. If the collection of last digits is random, then we would expect each integer to occur, on average, equally often. • Integer 0 1 2 3 4 5 6 7 8 9 . • Frequency 8 6 7 3 5 7 1 4 6 3 • Do the observed frequencies meet this expectation?

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