Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 1. Lecture Slides Elementary Statistics Eleventh Edition and the Triola.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 2. 11-1 Review and Preview 11-2 Goodness-of-fit 11-3 Contingency Tables 11-4 McNemar’s Test for Matched Pairs Chapter 11 Goodness-of-Fit and Contingency Tables

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 4. Preview  We focus on analysis of categorical (qualitative ) data that can be separated into different categories (often called cells).  Hypothesis test: Observed counts agree with some claimed distribution.  Also consider the contingency table or two-way frequency table (two or more rows and columns).  Two-way tables involving matched pairs.  Use the  2 (chi-square) distribution.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 7..  2 =  ( O – E ) 2 E Test Statistic: Rationale: -- measure of discrepancy between observed and expected frequencies. Simply summing the differences (O - E) is always = n – n = 0. Squaring avoids this problem (just like for standard deviation (x-xbar)^2). ∑ (O – E ) ^2 measures only the magnitude of differences, but we need to find magnitude of differences relative to what is expected. Therefore we divide by expected frequencies. The theoretical distribution of the right hand side sum is discrete because the number of possible values is finite. Chi-squared is an approximation, which is continuous. The approximation is acceptable as long as all expected values of E >= 5. If not, we have to combine some categories. The number of degrees of freedom = k – 1 reflects the fact that we can freely assign frequencies to (k – 1) categories before frequency for every category can be determined.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 8.. Expected Frequencies If all expected frequencies are equal: the sum of all observed frequencies divided by the number of categories n E = k

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 9.. If expected frequencies are not all equal: Each expected frequency is found by multiplying the sum of all observed frequencies by the probability for the category. E = np Expected Frequencies

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 10. NOTE: Observed frequencies must all be whole numbers (like the actual roll of a die). Expected values need not be whole numbers (expected value for a fair die is 1*(1/6)+2*(1/6) + … = 33/6 = 5.5 – not whole number). We know that sample frequencies always deviate somewhat from the theoretically expected values, which leads to the key question: Are the differences between the actual observed values O and theoretically expected values E statistically significant??? We need to measure the discrepancy between corresponding O and E values, so we use the test statistic measuring such relative discrepancy:  2 = ∑ (O-E)^2/E.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 11..  A large disagreement between observed and expected values will lead to a large value of  2 and a small P -value.  A significantly large value of  2 will cause a rejection of the null hypothesis of no difference between the observed and the expected.  A close agreement between observed and expected values will lead to a small value of  2 and a large P -value. Goodness-of-Fit Test

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 12.. Relationships Among the  2 Test Statistic, P-Value, and Goodness-of-Fit Figure 11-2 H0: no difference between the observed and the expected H1: there is difference

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 13.. Example: Data Set 1 in Appendix B includes weights from 40 randomly selected adult males and 40 randomly selected adult females. Those weights were obtained as part of the National Health Examination Survey. When obtaining weights of subjects, it is extremely important to actually weigh individuals instead of asking them to report their weights. By analyzing the last digits of weights, researchers can verify that weights were obtained through actual measurements instead of being reported.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 14.. Example: When people report weights, they typically round to a whole number, so reported weights tend to have many last digits consisting of 0. In contrast, if people are actually weighed with a scale having precision to the nearest 0.1 pound, the weights tend to have last digits that are uniformly distributed, with 0, 1, 2, …, 9 all occurring with roughly the same frequencies. Table 11-2 shows the frequency distribution of the last digits from 80 weights listed in Data Set 1 in Appendix B. For example, the weight of 201.5 lb has a last digit of 5, and this is one of the data values included in Table 11-2.) Test the claim that the sample is from a population of weights in which the last digits do not occur with the same frequency. Based on the results, what can we conclude about the procedure used to obtain the weights?

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 15.. Requirements are satisfied: 1) randomly selected subjects 2) data consists of frequency counts 3) with 80 sample values and 10 categories that are claimed to be equally likely, each expected frequency is 80/10 = 8 > 5 Step 1:The original claim is that the digits do NOT occur with the same frequency: at least one of the probabilities p 0, p 1,… p 9, is different from the others Step 2:If the original claim is false, then all the probabilities are the same. That is:p 0 = p 1 = p 2 = p 3 = p 4 = p 5 = p 6 = p 7 = p 8 = p 9 Step 3:null hypothesis contains equality H 0 : p 0 = p 1 = p 2 = p 3 = p 4 = p 5 = p 6 = p 7 = p 8 = p 9 H 1 : At least one probability is different

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 16.. Example: Step 4: no significance specified, so we just select to use  = 0.05 Step 5: testing whether a uniform distribution so use goodness-of-fit test:  2 Step 6: see the next slide for the computation of the  2 test statistic. The test statistic  2 = 11.250, using  = 0.05 and k – 1 = 9 degrees of freedom, the critical value is  2 = 16.919 STATDISK Enter observed frequencies into 1 st column of the Data Window If the expected frequencies are not all equal, enter 2 nd column of expected proportions or actual frequencies. Analysis -> Goodness Of Fit Num Categories: 10 Degrees of freedom: 9 Expected Freq: 8 Test Statistic, X^2: 11.2500 Critical X^2: 16.91895 P-Value: 0.2590 Fail to Reject the Null Hypothesis Data does not provide enough evidence to conclude that the proportions are not as claimed. Step 7: Because the test statistic does not fall in the critical region, there is not sufficient evidence to reject the null hypothesis. TI 84 Stat->Tests->Goodness of fit

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 17.. Example: Step 8: There is not sufficient evidence to support the claim that the last digits do not occur with the same relative frequency. This goodness-of-fit test suggests that the last digits provide a reasonably good fit with the claimed distribution of equally likely frequencies. Instead of asking the subjects how much they weigh, it appears that their weights were actually measured as they should have been.  # H0: Observed values are NOT significantly different from expected ones  # H1: Observed values are significantly different from expected ones > > n = 80; # number of sample points > k = 10; # number of categories > O = c(7,14,6,10,8,4,5,6,12,8) > E = rep(n/k,k) > E [1] 8 8 8 8 8 8 8 8 8 8 > > chisq = sum( (O-E)^2/E ) > chisq [1] 11.25 > > alpha = 0.05 > chisq_crit = qchisq(1-alpha,df = k-1) > chisq_crit [1] 16.91898 > # Classical Approach: chisq > # NOT reject H0, NOT support H1 > pvalue = 1 - pchisq(chisq,df=k-1) > pvalue # p_value is NOT small => SAME conclusion [1] 0.2589613

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 19. STATDISK Num Categories: 4 Degrees of freedom: 3 Test Statistic, X^2: 7.8848 Critical X^2: 7.814736 P-Value: 0.0485 Reject the Null Hypothesis Data provides evidence that the populations have proportions different from the expected proportions

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 21.  # H0: Observed values are NOT significantly  different from expected ones  # H1: Observed values are significantly  different from expected ones > > O = c(19,21,22,37) > n = sum(O) > k = length(O) > P = c(2,4,5,5)/16 > E = n*P > E [1] 12.3750 24.7500 30.9375 30.9375 > > chisq = sum( (O-E)^2/E ) > chisq [1] 7.884848 > > alpha = 0.05 > chisq_crit = qchisq(1-alpha,df = k-1) > chisq_crit [1] 7.814728 > # Classical Approach: : chisq > chisq_crit => > # reject H0, support H1 > > pvalue = 1 - pchisq(chisq,df=k-1) > pvalue # p_value is small => SAME conclusion [1] 0.04845246 > plot(x,O/n,"b") > lines(x,P,"b")

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 22. Num Categories: 4 Degrees of freedom: 3 Test Statistic, X^2: 119.4419 Critical X^2: 7.814736 P-Value: 0.0000 Reject the Null Hypothesis Data provides evidence that the populations have proportions different from the expected proportions

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 25. Num Categories: 6 Degrees of freedom: 5 Test Statistic, X^2: 4.7500 Critical X^2: 11.07051 P-Value: 0.4471 Fail to Reject the Null Hypothesis Data do not provide enough evidence to conclude that the populations have proportions different from the expected proportions

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 28.. Key Concept In this section we consider contingency tables (or two- way frequency tables), which include frequency counts for categorical data arranged in a table with at least two rows and at least two columns. We present a method for testing the claim that the row and column variables are independent of each other. We will use the same method for a test of homogeneity, whereby we test the claim that different populations have the same proportion of some characteristics.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 30.. A contingency table (or two-way frequency table) is a table in which frequencies correspond to two variables. (One variable is used to categorize rows, and a second variable is used to categorize columns.) Contingency tables have at least two rows and at least two columns. Example: Contingency table for Echinacea Treatment. r = 2 rows, c = 3 columns Cells contain frequencies. Row variable identifies if the subjects became infected. Column variable identifies the treatment group. Note that this is two-way frequency table: 1 st -way is row subjects infected or not 2 nd –way is column variable being one of the 3 options below. PlaceboEchinacea: 20% extract Echinacea: 60% extract Total Infected884842178 Not Infected1541029 Total10352 207

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 31.. Test of Independence A test of independence tests the null hypothesis that in a contingency table, the row and column variables are independent. This procedure cannot be used to establish a direct cause-and-effect link between variables in question. Dependence means only there is a relationship between the two variables.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 33. The test statistic allows us to measure the amount of disagreement between the observed and expected when two variables are actually independent. Large values of  2 test statistic are in the rightmost region of the chi-squared distribution, and they signal significant differences between observed and expected frequencies. The distribution of the test statistic is again discrete, but can be approximated by  2 distribution, provided that all expected frequencies >= 5. The number of degrees of freedom = (r-1)*(c-1) reflects the fact that because we know the total of all frequencies in a contingency table, we can freely assign frequencies to only (r-1) rows and (c-1) columns before the frequency for every cell is determined. PlaceboEchinacea: 20% extract Echinacea: 60% extract Total Infected884842178 Not Infected1541029 Total10352 207

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 34. PlaceboEchinacea: 20% extract Echinacea: 60% extract Total Infected178 Not Infected29 Total10352 207 To find expected frequencies, pretend that we know only the row and column totals, as in the table below, and we must fill in the cell expected frequencies by assuming independence (no relationship) between row and column variables. In the 1 st row, 178 of the 207 subjects got infections => P(infection) = 178/207 In the 1 st column, 103 of the 207 subjects were given a placebo => P(placebo) = 103/207 Because we assumed independence between getting infection and the treatment group (i.e. the treatment does not do anything) P(A and B) = [ A and B are independent ] = P(A)*P(B) => P(infection and placebo) = [ infection and placebo are independent ] = P(infection)*P(placebo) = (178/207) * (103/207) We can now find the expected value for the 1 st cell (infection and placebo) by multiplying the probability for that cell the total number of subjects: E = n*p = 207 * (178/207) * (103/207) = (178*103) / 207 = 88.57 Generalizing we get on the next slide: Back to Echinacea Example

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 35... Expected Frequency for Contingency Tables E = grand total row total column total grand total E = (row total) (column total) (grand total) (probability of a cell) n p

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 36.. Example: Common colds are typically caused by a rhinovirus. In a test of the effectiveness of echinacea, some test subjects were treated with echinacea extracted with 20% ethanol, some were treated with echinacea extracted with 60% ethanol, and others were given a placebo. All of the test subjects were then exposed to rhinovirus. Results are summarized in Table 11-6. Use a 0.05 significance level to test the claim that getting an infection (cold) is independent of the treatment group. What does the result indicated about the effectiveness of echinacea as a treatment for colds?

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 37.. Example: Requirements are satisfied: 1) the subjects are randomly assigned to treatment groups 2) the results are expressed as frequency counts 3) expected frequencies are all >= 5 (will see later) H 0 :Getting an infection is independent of the treatment H 1 :Getting an infection and the treatment are dependent PlaceboEchinacea: 20% extract Echinacea: 60% extract Total Infected884842178 Not Infected1541029 Total10352 207

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 38.. Book approach: The critical value of  2 = 5.991 is found from Table A-4 with  = 0.05 in the right tail and the number of degrees of freedom given by (r – 1)(c – 1) = (2 – 1)(3 – 1) = 2. Because the test statistic does not fall within the critical region, we fail to reject the null hypothesis of independence between getting an infection and treatment. STATDISK Enter observed contingencies in the data window. Analysis->Contingency Tables Degrees of freedom: 2 Test Statistic, X^2: 2.9255 Critical X^2: 5.991471 P-Value: 0.2316 Fail to Reject the Null Hypothesis Data does not provide enough evidence to conclude that the rows and columns are related My way: P-value = 0.23 is large => Not enough evidence to reject H0 that they are independent It appears that getting an infection is independent of the treatment group. This suggests that echinacea is not an effective treatment for colds.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 41. Example 4: Is nurse a serial killer? (p 585) Test a claim that whether nurse Gilbert was on duty is independent of whether a patient died during the shift. Use significance level alpha = 0.01 Requirements 1) Treat the sample data as random 2) Frequency counts in two way table – yes. 3) Each expected frequency >= 5 H0: Gilbert working is independent of death H1: --- dependent There is sufficient evidence to warrant rejection of independence claim. It appears that there is an association. Note that it does not prove that she caused death. => not evidence of wrong doing for a trial. Shifts with death Shifts without death Gilbert working 40217 Gilbert not working 341350 Degrees of freedom: 1 Test Statistic, X^2: 86.4809 Critical X^2: 3.841456 P-Value: 0.0000 Reject the Null Hypothesis Data provides evidence that the rows and columns are related

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 43. Child buckledChild unbuckled Parent buckled568 Parent unbuckled 216 My Example: Seat belt usage data. Test a claim that whether parent is buckled is independent of whether a child is buckled. Use significance level alpha = 0.01 Requirements 1) Treat the sample data as random 2) Frequency counts in two way table – yes. 3) Each expected frequency >= 5 H0: independent H1: dependent There is sufficient evidence to warrant rejection of independence claim. It appears that there is an association. Degrees of freedom: 1 Test Statistic, X^2: 39.5993 Critical X^2: 3.841456 P-Value: 0.0000 Reject the Null Hypothesis Data provides evidence that the rows and columns are related

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 45.. Part 2:Test of Homogeneity and the Fisher Exact Test In part 1, we focused on the test of independence between the row and column variables in a contingency table. The sample data were from one population, and individual sample results are categorized with the row and column variables. Here, we obtain samples from different populations, and we want to determine if those populations have the same proportions of the characteristics being considered. Test of Homogeneity In a test of homogeneity, we test the claim that different populations have the same proportions of some characteristics.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 46.. How to Distinguish Between a Test of Homogeneity and a Test for Independence For test of homogeneity - predetermined sample sizes are used for different populations. For test of independence - one big sample is drawn, so both row and column totals were determined randomly (test of independence)? Homogeneity test used the same notation, requirements, test statistic, critical values and procedures used in Part 1 for independence test. The only difference is that instead of testing null hypothesis of independence between the row and column variables, we test the null hypothesis the different populations have the same proportions for some statistics. I.e. the distributions are the same.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 47.. Does a pollster’s gender have an effect on poll responses by men? A U.S. News & World Report article about polls stated: “On sensitive issues, people tend to give ‘acceptable’ rather than honest responses. Their answers may depend on the gender or race of the interviewer.” To support that claim, data were provided for an Eagleton Institute poll in which surveyed men were asked if they agreed with this statement: “Abortion is a private matter that should be left to the woman to decide without government intervention.” Example:

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 48.. We will analyze the effect of gender on male survey subjects only. Table 11-8 is based on the responses of surveyed men. Assume that the survey was designed so that male interviewers were instructed to obtain 800 responses from male subjects, and female interviewers were instructed to obtain 400 responses from male subjects. Using a 0.05 significance level, test the claim that the proportions of agree/disagree responses are the same for the subjects interviewed by men and the subjects interviewed by women. 800 400 868 332 1200

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 49.. H 0 : The proportions of agree/disagree responses are the same (homogeneous) for the subjects interviewed by men and the subjects interviewed by women. H 1 : The proportions are different. Requirements are satisfied: data are random, frequency counts in a two-way table, expected frequencies are all at least 5 Example: Test for homogeneity.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 50.. STATDISK Degrees of freedom: 1 Test Statistic, X^2: 6.5293 Critical X^2: 3.841456 P-Value: 0.0106 Reject the Null Hypothesis Data provides evidence that the rows and columns are related The Minitab display (book) shows the expected frequencies of 578.67, 289.33, 221.33, and 110.67. It also includes the test statistic of  2 = 6.529 and the P-value of 0.011. Using the P-value approach to hypothesis testing, we reject the null hypothesis of equal (homogeneous) proportions (because the P-value of 0.011 is less than 0.05). There is sufficient evidence to warrant rejection of the claim that the proportions are the same. It appears that response and the gender of the interviewer are dependent. Although this statistical analysis cannot be used to justify any statement about causality, it does appear that men are influenced by the gender of the interviewer.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 52.. The procedures for testing hypotheses with contingency tables with two rows and two columns (2  2) have the requirement that every cell must have an expected frequency of at least 5. This requirement is necessary for the  2 distribution to be a suitable approximation to the exact distribution of the test statistic. When expected frequencies < 5 in a bigger table, one approach is to combine some categories, but for 2  2 table you cannot go any smaller. The Fisher exact test is often used for a 2  2 contingency table with one or more expected frequencies that are below 5. The Fisher exact test provides an exact P-value and does not require an approximation technique. Because the calculations are quite complex, it’s a good idea to use computer software when using the Fisher exact test. STATDISK and Minitab both have the ability to perform the Fisher exact test. Fisher Exact Test skip

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 54. McNemar’s Test uses frequency counts from matched pairs of nominal data from two categories to test the null hypothesis that for a 2  2 table such as Table 11-9, the frequencies b and c occur in the same proportion.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 1. Lecture Slides Elementary Statistics Eleventh Edition and the Triola.

Similar presentations

Presentation on theme: "Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 1. Lecture Slides Elementary Statistics Eleventh Edition and the Triola."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 1. Lecture Slides Elementary Statistics Eleventh Edition and the Triola.

Similar presentations

Presentation on theme: "Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 11.1 - 1. Lecture Slides Elementary Statistics Eleventh Edition and the Triola."— Presentation transcript:

Similar presentations

About project

Feedback