...The goal of this exercise is to assess your understanding of the chi square test. Please read the problem carefully and answer the question. Good luck! Assume you have data below that displays the number of students who elect different undergraduate majors. Number of Students Selecting Different Majors | |Computer Sciences |English Literature | | | | |Pre-Med | | |Education |Engineering |Total | |50 |85 |25 |60 |80 |300 | We want to know whether those numbers differ due to chance. In other words, at 0.01 level of confidence, are some majors selected more often than others, or is the selection pattern essentially random? The null hypothesis is that the programs are equally preferred. Create a table that shows the computation of the Chi Square statistic [6 POINTS]. Use a decision rule to determine whether the null hypothesis is rejected or not [4 POINTS]. Solution: Ho: The majors are equally preferred (probability of liking each major = 1/5). HA: The majors are not equally preferred. (Using the Chi Square Statistic to evaluate to what extent the hypothesis and data have a good fit. [pic] Where, Oi is actual frequency observed in cell i ...
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...0% | 0 | .0% | 300 | 100.0% | Sex * Chatting | 300 | 100.0% | 0 | .0% | 300 | 100.0% | Sex * News/Weather | 300 | 100.0% | 0 | .0% | 300 | 100.0% | Sex * Music Crosstab | | Music | Total | | 0 | 1 | | Sex | 1 | Count | 91 | 97 | 188 | | | % within Sex | 48.4% | 51.6% | 100.0% | | | % within Music | 65.0% | 60.6% | 62.7% | | 2 | Count | 49 | 63 | 112 | | | % within Sex | 43.8% | 56.3% | 100.0% | | | % within Music | 35.0% | 39.4% | 37.3% | Total | Count | 140 | 160 | 300 | | % within Sex | 46.7% | 53.3% | 100.0% | | % within Music | 100.0% | 100.0% | 100.0% | Chi-Square Tests | | Value | df | Asymp. Sig. (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | Pearson Chi-Square | .611a | 1 | .434 | | | Continuity Correctionb | .438 | 1 | .508 | | | Likelihood Ratio | .612 | 1 | .434 | | | Fisher's Exact Test | | | | .474 | .254 | Linear-by-Linear...
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...analysis a. For two interval/ratio variables – use correlation b. For two nominal/ordinal variables – use cross-tabs 4. Does a relationship exist? 5. If relationship exists, determine the direction a. Monotonic will be increasing/decreasing b. Nonmonotonic will be looking for a pattern 6. Assess the strength of relationship a. With correlation – size of coefficient denotes the strength b. With cross-tabs – the pattern is assessed Cross-Tabulations and Chi Square • Cross-tabulations o Consists of rows and columns defined by the categories classifying each variable. Used for nonmonotonic relationships o Sometimes referred to as an “r x c” table (rows x columns) ▪ Crosstabulation cell – intersection of a row and a column o Interested in inner cells to determine relationship before statistically testing ▪ Use the chi-square for statistical tests o Tables consist of four types of numbers in each cell: ▪ Frequency ▪ Raw percentage ▪ Column percentage ▪ Row percentage o When we have two nominal-scaled variables and we want to know if they are associated, we use cross-tabulations...
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...Lab assignment 2 Part 1: Correlation Analysis 1. Explain when we run a correlation analysis? (1 Mark) Answer: We run a correlation to see if there are any significant associations between variables and/or before we do a regression analysis. 2. Interpret the correlations. Which variables are correlated? (hint: as p/sig values are provided use a cut-off of 0.05 to determine significance). Explain the nature of each relationship (8 Marks) Answer: a) Inquiry and Memory variables are significantly correlated as suggested by their Sig value of .000 (p < .001) which is less than 0.05. The Pearson Correlation of .718 indicates a strong positive correlation. b) Satisfaction with rates and Memory variables are significantly correlated as suggest by their Sig value of 0.010 (p < .05) which is below 0.05. The Pearson Correlation of -.563 indicates a moderate negative correlation. c) Satisfaction with service and Service perception variables are significantly correlated as suggest by their Sig value of 0.000 (p < .001) which is well below 0.05. The Pearson Correlation of .710 indicates a strong positive correlation. d) Memory and Rates perception variables are significantly correlated as suggest by their Sig value of 0.010 (p < .05) which is below 0.05.The Pearson Correlation of .560 indicates a moderate positive correlation. 3. What are your recommendations to the manager (refer back to the decision problem) (3 Marks) Answer: Sun Insurance...
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...Pearson Chi-Square significance value is less than .05 which means income can affect the probability that a person will eat at Hobbit’s Choice. Probable Hobbit’s patrons are more likely to make between $50,000 and 74,999 (93%) a year than non-probable patrons (7%). Income | Probable Patron | Non-Probable Patron | <$15,000 | 0% | 100% | $15,000 to 24,999 | 0% | 100% | $25,000 to 49,999 | 0% | 100% | $50,000 to 74,999 | 3% | 97% | $75,000 to 99,999 | 62.5% | 32.5% | $100,000 to 149,999 | 93% | 7% | $150,000+ | 84.8% | 15.2% | *Please see Appendix ____ for SPSS Output * The Pearson Chi-Square significance value is less than .05 which means that educational level has an effect on the probability that a person will be a patron of Hobbit’s Choice. In other words, level of education differentiates patrons from non-patrons. Probable Hobbit’s Choice patrons are more likely to have a Doctorate degree (77.8%) than non-patrons (22.2%). In fact, most/all (which one?) probable patrons have more than some college. 0% of survey respondents that list “no degree” are probable patrons. Educational Level | Probable Patron | Non-Probable Patron | Some College or Less | 0% | 100% | Associate Degree | 21.4% | 78.6% | Bachelor’s Degree | 27.7% | 72.3% | Master’s Degree | 39.5% | 60.5% | Doctorate Degree | 77.8% | 22.2% | *Please see Appendix ____ for SPSS Output * Gender does not differentiate patrons from non-patrons because its Pearson Chi-Square significance...
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...A STUDY ON TRAFFIC RULES VIOLATIONS IN CHENNAI (VANDALUR AND PERANGALUTUR) Submitted By: Praveen Kumar A, MBA. INTRODUCTION Violations in traffic laws are very common in a highly populated country like India. The conditions are even worse in metro cities like Delhi, Mumbai Bangalore and Chennai. The accidents associated with these violations cause a huge loss to life and property. Same is the case in Chennai. Being a metro city and a highly populated one also, has a lot of road accidents every year. Despite this the violations in traffic laws do not reduce. A lot of people disobey the rules every day sometimes willingly and sometimes because they are forced to do so because of others. The major traffic laws in India are wearing a helmet in case of two wheelers, putting a seat belt in case of cars, driving on the right side of the road including overtaking from the right direction, over speeding in certain restricted areas, not obeying the traffics signals and stooping the car after the finish line. It is mainly because of these violations that major accidents occur. It should be recognized that the highway is a social situation, in which people are interacting. However the drivers are unknown to each other in most of the cases and the interaction between them very brief and non-recurring. The communication between them is very limited and that also through mechanical aids like lights and horns. The main objective of these laws is to minimize the confusion...
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...Abstract The “Brassica rapa” is a fast plant known as the field mustard. This plant is well known for its rapid growing rate, which makes it an easy breeding cycle and easy to pollinate. In giving so this makes “Brassica rapa” a great participant for testing Gregor Mendel’s theories of inheritance. The “Brassica rapa” acts like a test subject in testing cross-pollination giving the understanding to the dominant allele of colored stems. There are different colors that are visible on the stem that are above the soil; the colors vary from green to purple. P1 seed was ordered, germinated and cross-pollinated until germination of the next off spring of plants were also done. It was hypothesized that the presence of purple pigment (anthocyanin) is inherited as a dominant trait and follows Mendelian laws. The experiments were performed in quads, with the different quads being amongst different sections; those with data were then gathered. The complete data was used to prove or disagree with Mendel’s theories. The data was to represent which of he two colors were a dominating phenotype. At the end of the experiment, realization of errors came to show by the Chi Square, results were considered questionable. Materials The following materials are present to correctly repeat this experiment. These materials are: seeds of specific phenotypes, seed-collecting pan, small envelopes, several wicks, opaque tape, Styrofoam quads, fluorescent light bank, water reservoir, water, water dropper, potting...
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...CHI-SQUARE TEST Adapted by Anne F. Maben from "Statistics for the Social Sciences" by Vicki Sharp The chi-square (I) test is used to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. Do the number of individuals or objects that fall in each category differ significantly from the number you would expect? Is this difference between the expected and observed due to sampling error, or is it a real difference? Chi-Square Test Requirements 1. Quantitative data. 2. One or more categories. 3. Independent observations. 4. Adequate sample size (at least 10). 5. Simple random sample. 6. Data in frequency form. 7. All observations must be used. Expected Frequencies When you find the value for chi square, you determine whether the observed frequencies differ significantly from the expected frequencies. You find the expected frequencies for chi square in three ways: I . You hypothesize that all the frequencies are equal in each category. For example, you might expect that half of the entering freshmen class of 200 at Tech College will be identified as women and half as men. You figure the expected frequency by dividing the number in the sample by the number of categories. In this exam pie, where there are 200 entering freshmen and two categories, male and female, you divide your sample of 200 by 2, the number of categories, to get 100 (expected frequencies) in each category. 2. You determine the expected...
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...CHI-SQUARE TEST - ANALYSIS OF CONTINGENCY TABLES David C. Howell University of Vermont The term ”chi-square” refers both to a statistical distribution and to a hypothesis testing procedure that produces a statistic that is approximately distributed as the chi-square distribution. In this entry the term is used in its second sense. PEARSON’S CHI-SQUARE The original chi-square test, often known as Pearson’s chi-square, dates from papers by Karl Pearson in the earlier 1900s. The test serves both as a ”goodnessof-fit” test, where the data are categorized along one dimension, and as a test for the more common ”contingency table”, in which categorization is across two or more dimensions. Voinov and Nikulin, this volume, discuss the controversy over the correct form for the goodness of fit test. This entry will focus on the lack of agreement about tests on contingency tables. In 2000 the Vermont State legislature approved a bill authorizing civil unions. The vote can be broken down by gender to produce the following table, with the expected frequencies given in parentheses. The expected frequencies are computed as Ri × Cj /N, where Ri and Cj represent row and column marginal totals and N is the grand total. Vote Women Men Total Yes 35 (28.83) 60 (66.17) 95 No 9 (15.17) 41 (34.83) 50 Total 44 101 145 The standard Pearson chi-square statistic is defined as χ2 = (Oij − Eij )2 (35 − 28.83)2 (41 − 34.83)2 = + ··· + = 5.50 Eij 28.83 34.83 where i and j index the rows and columns of...
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...Case 14.1 1. Correlations | | Prefer Drive Less than 30 Minutes | Prefer Unusual Desserts | Prefer Large Variety of Entrees | Prefer Unusual Entrees | Prefer Drive Less than 30 Minutes | Pearson Correlation | 1 | .768** | .806** | .765** | | Sig. (2-tailed) | | .000 | .000 | .000 | | N | 400 | 400 | 400 | 400 | Prefer Unusual Desserts | Pearson Correlation | .768** | 1 | .823** | .868** | | Sig. (2-tailed) | .000 | | .000 | .000 | | N | 400 | 400 | 400 | 400 | Prefer Large Variety of Entrees | Pearson Correlation | .806** | .823** | 1 | .831** | | Sig. (2-tailed) | .000 | .000 | | .000 | | N | 400 | 400 | 400 | 400 | Prefer Unusual Entrees | Pearson Correlation | .765** | .868** | .831** | 1 | | Sig. (2-tailed) | .000 | .000 | .000 | | | N | 400 | 400 | 400 | 400 | **. Correlation is significant at the 0.01 level (2-tailed). | Null Hypothesis- No relation between preference to drive 30 minutes or less and preference of menu items Alternative Hypothesis- There is a relation between the preference to drive 30 minutes or less and preference of menu items Interpretation-All the correlations have sig values that are significantly different from zero. So, we reject the null hypothesis. The correlations are positive and they are in the moderate range. As the preference to drive 30 minutes or less increases, so do preferences for unusual deserts, large variety of entrees, and unusual entrees. Correlations | | Prefer Drive Less than 30 Minutes | Prefer...
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...Quiz 11 1）The test statistics for a small-sample test concerning proportion is the _____（hypothesis）______. [hint: use all lowercase letters] (wrong)try hypotheses (wrong) 2) The chi-square distribution is symmetrical. false (right) True False 3) If the observed frequencies are exactly equal to the expected frequencies in a chi-square test, the value of χ2 is: (0)(right) | | close to 1 | | | a large positive value | | | 0 | | | a very small positive value | 4) The chi-square test can be used to determine if there is a significant difference between two sample proportions.(true)(right)True False 5) The null hypothesis that two processes produce the same proportion of defectives can be written: (p1-p2=0)(right) | | p = 0 | | | x1/n1 = x2/n2 | | | p1 - p2 = 0 | | | x/n = 0 | 6） Which of the following null hypotheses cannot be tested with a chi-squared test? (last one) right | | Ho: the two variables defining the contingency table are independent. | | | Ho: the population distribution is a normal distribution. | | | Ho: the true category proportions are the same for all of the population. | | | Ho: the true population means are the same for all of the populations | | | | 7) The larger the calculated χ2 value in a goodness-of-fit test, the _______ likely the data fits the hypothesized distribution.( chi-square)(wrong)try observed (wrong) try less8) A chi-square statistic can be used...
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...interior design of restaurants. Previous research conducted on consumers has shown that location has been a factor influencing a consumer’s decision to go to a restaurant. It has also shown that locational variables such as environment and packet quality have been considered important to consumers. Consumer demographics if affects the consumer perception then it may affect the consumer’s decision to go to a restaurant. The correlation between different locational variables we found out by calculating Pearson’s correlation coefficient and effect of the demographic features on consumer preference was found out through cross tabulation. The results were tested for significance by calculating probabilities (2 tailed) and by performing a chi squared test respectively. ------------------------------------------------- Importance For the Owner: With increasing popularity of restaurants, more and more consumers are going to restaurants in order to eat. Considering the large population of Dhaka which is currently15, 414,000 (Cox 2012)...
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...Effects of Bullying among Children and Adults Jose Vargas PSY325 Prof. Mar Navarro Submitted 5/7/2012 Currently, bullying is a large problem in schools which causes significant stresses to its victims (Voss & Mulligan, 2000). Bullying became a greater concern for school personnel, parents & research after a series of school shootings in the late 1990s, including Columbine (Seals & Young, 2003). The effects of bullying are seen over a long period of time in the lives of participants and can lead to antisocial behavior amongst both perpetrators and victims of bullying (Voss & Mulligan, 2000; Seals & Young, 2003.) This paper intends to conduct an overview on the effects that bullying has on its victims, both adult and child and which group is more or less likely to become bullied. It is also important to examine how the effects of bullying differ by age, gender, ethnicity and other factors that lower social status which predispose people to bullying. This paper proposes that bullying impacts groups differently and thus the symptomatology that results will differ, with those who are socially disadvantaged experiencing the greatest impact of bullying on their lives. Bullying should also show strong outcomes for negative social adjustment in its aftermath, including increased depression, stress, alcohol abuse and decreased self-esteem. Bullying is defined as repeated aggressive behavior which is intended to harm or disturb a person in which the conflict is...
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...Chapter 23 Chi-Square Tests 23 Chi-Square Procedures The Chi-Square Formula The Chi-Square Critical Value Chi-Square Goodness of Fit Test Chi-Square Test of Independence Cautions in Using Chi-Square Dr. Helen Ang studied the relationship between predominant leadership style and educational philosophy of administrators in Christian colleges and universities for her Ed.D. dissertation in 1984.1 Leadership Style was a categorical variable with the following five levels (with percentages of the 113 administrators studied): team administrator (high people/high task: 23%), constituency-centered (moderate people/moderate task: 16%), authorityobedience (low people/high task: 4%), comfortable-pleasant (high people/low task: 38%), and caretaker (low people/low task: 19%).2 Educational Philosophy Profile was a categorical variable with the following six levels (with percentages): idealism (7%), realism (4%), neo-thomism (15%), pragmatism (58%), existentialism (1%), and “eclectic” (16%).3 Applying the Chi-Square Test of Independence, Dr. Ang found that the variables Leadership Style and Educational Philosophy were independent (χ2 = 21.676, χ2cv = 31.410, a=0.05, df=20).4 The chi in chi-square is the Greek letter χ, pronounced ki as in kite. Chi-square (χ2) procedures measures the differences between observed (O) and expected (E) frequencies of nominal variables, in which subjects are grouped in categories or cells. There are two basic types of chi-square analysis,...
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...Table of Content Acknowledgement: 3 Executive summary: 4 Methodology 7 Sampling Design 8 Limitation of the report 9 Analysis of finding 10 Terminology 10 Findings 11 SQ 7 Caste Ethnicity * Q5 Support Fairly 3 SQ 8 Occupation * Q5 Support Fairly 6 SQ 8 Occupation * Q8 Post Earth quake 8 SQ 5 Age * Q8 Post Earth quake 10 SQ 5 Age * Q5 Support Fairly 11 SQ 6 Gender * Q1 Problem Addressed 13 Conclusion: 16 Annex 17 SQ 8 Occupation * Q1 Problem Addressed 17 SQ5Age * Q2Satisfied 18 SQ6Gender * Q2Satisfied 19 SQ7CasteEthnicity * Q2Satisfied 20 SQ8Occupation * Q2Satisfied 21 SQ5Age * Q3InfoReliefAndSupport 23 SQ6Gender * Q3InfoReliefAndSupport 24 SQ7CasteEthnicity * Q3InfoReliefAndSupport 25 SQ8Occupation * Q3InfoReliefAndSupport 26 SQ5Age * Q4NGoWork 28 SQ6Gender * Q4NGoWork 29 SQ7CasteEthnicity * Q4NGoWork 30 SQ8Occupation * Q4NGoWork 32 SQ6Gender * Q5SupportFairly 33 SQ 5 Age * Q6 Prepared For Winter 34 SQ 6 Gender * Q6 Prepared For Winter 35 SQ 7 Caste Ethnicity * Q6 Prepared For Winter 37 SQ 8 Occupation * Q6 Prepared For Winter 38 SQ 5 Age * Q7 Problem rel Children 39 SQ 6 Gender * Q7 Problem rel Children 41 SQ 7 Caste Ethnicity * Q7 Problem rel Children 42 SQ 8 Occupation * Q7 Problem rel Children 43 SQ 7 Caste Ethnicity * Q8 Post Earth quake 44 SQ 5 Age * Q9 Increase Voilence 45 SQ 6 Gender * Q9 Increase Voilence 46 SQ 7 Caste Ethnicity * Q9 Increase Violence 47 SQ 8 Occupation * Q9 Increase...
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