c) 7 d) 12 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence‚ degrees of freedom 2. If we use the [pic] method of analysis to test for the differences among 4 proportions‚ the degrees of freedom are equal to: a) 3. b) 4. c) 5. d) 1. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions‚ degrees of freedom 3. If we wish to determine
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CUSTOMER SATISFACTION IN HOTELS IN CAPE TOWN MBUNGWANA CHRISTINE LUNGISWA 2009 CUSTOMER SATISFACTION IN HOTELS IN CAPE TOWN by MBUNGWANA CHRISTINE LUNGISWA Dissertation submitted in fulfilment of the requirements for the degree Master of Technology: Quality in the Faculty of Engineering at the Cape Peninsula University of Technology Supervisor: A Bester Co-supervisor: Prof. Dr. J A Watkins D. Phil.‚ D. Com.‚ Ph. D. Bellville November 2009 ii DECLARATION I‚ Christine Mbungwana‚ hereby
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Crime Rates: An Econometric Analysis using population‚ unemployment and growth Table of Contents I. Introduction A.) Background of the Study B.) Problem Statement C.) Objectives D.) Significance of the Study E.) Scope and Limitations II. Review of Related Literature III. Operational Framework A.) Variable List B.) Model Specification C.) A-priori Expectations IV. Methodology A.) Data B.) Preliminary Tests V. Results and Discussions VI. Conclusion
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User Satisfaction Analysis- Facebook Contents Abstract This study focuses on analysing the impact of changes brought in Facebook interface and new services added on various categories of Facebook users. In order to analyse the impact a primary research has been conducted and various statistical tools have been used. 1 Introduction World’s largest social networking site FACEBOOK‚ with around 900 million users connecting with their friends‚ families‚ work
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Statistics Cheat Sheet Proportion = Frequency x 100 = Percentage Total No | Z score (standardised value)-how many sds from the mean the value liesZ score = data value – mean Standard deviation | Metric Data = ExploreCategory = Frequencies | Bigger sample size will give a narrower confidence interval range (more specific) outliers affect the mean but not the median – this is why the median is preferred here.mean | | Reports -Only give confidence
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Determining Allele Frequencies of the PV92 Alu Element using DNA Isolated from Human Cheek Cells and PCR Amplification Background Alu elements are the most abundant repetitive elements in the human genome that have mobilized throughout primate genomes by retrotransposition over the past 65 million years ago from a 5’ to 3’ fusion of the 7SL RNA gene‚ to reach the present number of more than one million copies. Over the last few years‚ several lines of evidence demonstrated that these elements
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Name __________________________________ AP Biology Period _________ Date ______________________ AP: LAB-RELATED AP EXAM ESSAYS LAB 1. OSMOSIS AND DIFFUSION ESSAY 1992 A laboratory assistant prepared solutions of 0.8 M‚ 0.6 M‚ 0.4 M‚ and 0.2 M sucrose‚ but forgot to label them. After realizing the error‚ the assistant randomly labeled the flasks containing these four unknown solutions as flask A‚ flask B‚ flask C‚ and flask D. Design an experiment‚ based on the principles of diffusion
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Social Media for Business-An Analysis of Statistic Data Introduction. The area under discussion in the following assignment is to quantify the possibility of a deviation from accuracy in the fifty samples formed from a selection of data‚ following their questionnaire rejoinders. Questioning in which there is a relationship between the kinds of social media is and the type of business. Moreover‚ does this relationship exist for the reason that day by day more people use social media? (NielsenWire
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* Formulate an analysis plan. For this analysis‚ the significance level is 0.05. Using sample data‚ we will conduct a chi-square goodness of fit test of the null hypothesis. * Analyze sample data. Applying the chi-square goodness of fit test to sample data‚ we compute the degrees of freedom‚ the expected frequency counts‚ and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom‚ we determine the P-value. DF = k - 1 = 3 - 1 = 2 (Ei) = n * pi (E1) =
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express a desire to go to university and this would be the same across both genders. This data is represented in table ii. CHI-SQUARE EXPECTED FREQUENCIES Male [1] Yes [1] No [2] Total Table ii. If we now look at the actual responses given in table i we can see that the total responses for each case is somewhat different from what was expected. This is called the observed frequencies. This data is tabulated in table iii below. 20 10 30 Female [2] 20 10
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