...How to Use and Apply SPSS Version 15.0 for Windows Notes This mini guide/ walkthrough is for SPSS 15.0 for Windows and was written in the Summer of 2010. If there are any changes/updates that apply afterwards, this document will not address them. Additionally, if a previous version (pre 15.0) is being used, this document might not fully apply. This said, the concepts behind what is being explained should still be the same, so long as the functions of the program remain similar. This guide is intended to give the reader a VERY basic understanding on how to use SPSS. This is also intended to be a crash course type of guide. The length of this document is indicative of how “in depth” this document goes. Furthermore, there is a lot that this document does not mention. If you want to perform a very thorough analysis with very in depth statistics, you can read the SPSS survival manual by Julie Pallant. Table of Contents * Preparations3 * Getting Started4 * Entering Data7 * Output Window9 * Walkthrough10 * Analysis10 * Graphing10 * Regression13 * Correlation14 * Testing15 * One Sample T-Test 15 * Independent Sample T-Test 17 * Paired Sample T-Test 18 * One Way ANOVA 20 * Hypothesis Testing Crash Course26 Preparations * Know that SPSS is for analyzing data from a statistical researcher’s point of view. The name SPSS stands for Statistical Package for the Social Sciences – SPSS. This means that this program is best...
Words: 4680 - Pages: 19
...Part I. Data Description. 1.1. Original Data. Due to the recent restructure of the company that resulted in personnel redundancy, the top management of JSC “TESS” * requested to investigate the average level of the job satisfaction (JSL) of the employees. To carry out the analysis we have collected 203 questionnaires (Appendix _) as a sample of total population. Total population: 1200 employees. The main objectives of the analysis are: * To compare the mean JSL with the top-management statement; * To find out if there is a significant difference in mean JSL between the groups of employees with different work experience at the company. * To investigate if the mean JSL differs between the branches of the company. The data set used for the analysis: Variable | How the variable is measured | Branch | Branches of the company:1= TESS-Nizhnevartovsk, TESS-Kogalym2= TESS Head Office, TESS-Surgut3=TESS-Tyumen, TESS-Khanty-Mansiysk | Number | Number of the respondent | Work_Exp | Work Experience in JSC “TESS”:1= 2 year or less 2= more than 2 years | JSL | Job Satisfaction Level:Ratings from 1 to 5 where 1= very unsatisfied, 5= very satisfied and 0= no answer/blank | 1.2. Revised Data. Test for Normal Distribution To proceed with the analysis it is necessary to determine if the data are distributed normally. The Histogram below as well as the Descriptive Statistics (Appendix 1, Table 1b) show that the data distribution is leptokurtic (kurtosis...
Words: 2253 - Pages: 10
...Explore |Notes | |Output Created |12-Feb-2012 14:20:11 | |Comments | | |Input |Active Dataset |DataSet1 | | |Filter | | | |Weight | | | |Split File | | | |N of Rows in Working Data File |30 | |Missing Value Handling |Definition of Missing |User-defined missing values for dependent | | | |variables are treated as missing. | | |Cases Used |Statistics are based on cases with no missing | | | |values for any dependent variable or...
Words: 695 - Pages: 3
...8.46.) A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very accurate scale. The results in grams were 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477 (a) Construct a 90 percent confidence interval for the true mean weight. E = 1.96(s/sqrt(n)) = 1.96[0.131989/sqrt(10)]=1.96*0.41739 =0.081808 C.I. = (x-bar-E,x-bar+E) = (3.3048-0.0818,3.3048+0.0818) (b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? n=[z'*s/E]^2 n=[1.645*0.131989/0.03]^2 = 52.38; rounding up, n=53 (c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture. The equipment that the Tootsie Rolls are made on could be a factor if they are not calibrated properly or as intended. The volume of the equipment or the speed of the candy that is fed into the cutter also influences the weight. There could also be a weight variation if there is a fluctuation in the temperature. 8.62.) In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a 95 percent confidence interval for the population proportion of positive drug tests. E = 1.96*sqrt [0.01314*(1-0.01314)/86,991] = 0.0007567.. 95% CI: 0.79615, 83671 (b) Why is the normality assumption not a problem, despite the very small value of p? The normality...
Words: 326 - Pages: 2
...Exercises From the E-Text 5. State the main points of the Central Limit Theorem for a mean? “The Central Limit Theorem (CLT) is a powerful result that allows us to approximate the shape of the sampling distribution of X even when we don’t know the shape of the population distribution” (Doane & Seward, 2007, p. 299). 6. Why is population shape of concern when estimating a mean ? What does sample size have to do with it? The shape of a population is of concern when estimating a mean because the shape can determine if a population is normal thus providing the normality of the distribution of the sample mean. According to Applied Statistics in business and economics, “If the population is normal, the distribution of the sample mean is normal regardless of the sample size” (Doane & Seward, 2007, p. 299). | | Exercise 8.46 A random sample of 10 min tootsie rolls was taken from a bag. Each piece was weighted on a very accurate scale. The results in grams were: 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477 A) Construct a 90% confidence interval for the true mean weight Assuming the population has a normal distribution, we should use the Student t distribution E= z(sigma)/ sqrt(n) E= 1.645(0.13199/ sqrt(10) E= 0.06866 x-bar= 3.3048 90% CI = 3.3048-.06866= 3.23614, 3.3048 + 0.06866 = 3.37346 Confidence interval: 3.23614 to 3.37346 B) What sample size would be necessary to estimate the true weight with an error...
Words: 486 - Pages: 2
...nt Team Reflection Week 4 Team D Reflection Christopher Gonzalez, Jennifer Mai, Leonel Perez QNT 351 March 7, 2014 Mr. Victor Ornelas Week 4 Team D Reflection In week four, Team D read about the steps in testing a research hypothesis, comparing the means of two or more groups, and calculating the correlation between two variables. The following paper will go in to more depth on what was discussed and what we learned as a team as well as individuals. The team also was able to go over the Mystat lab that we were assigned in the individual assignment. In step one a team would state a null hypothesis as well as an alternate hypothesis. The null hypothesis is used only for testing. We either reject or fail to reject the null hypothesis. The alternate hypothesis is accepted if sample data provide enough evidence that the null hypothesis is false. No matter what the problem looks like the null hypothesis will always contain the equal sign. This is true because since the null hypothesis is being tested there is a need for a specific value to be seen in the calculations. When comparing the means of two or more groups the team must state the null hypothesis. Random samples of each population must be chosen to show the formula in a working status. To get a more accurate sample a higher number of the populations should be used. An independent sample T-test is used to compare the means of two or more groups. Pooled variance can be used if the variances of the two populations...
Words: 352 - Pages: 2
...Module 16 quiz * Question 1 10 out of 10 points | | | It is a well known fact that talking on a cell phone impairs a driver more than talking to a person sitting in the passenger seat. This is because a passenger can see the road and will stop talking during dangerous driving conditions. We want to know if this is only true for passengers that have experience driving. We recruit 40 participants to be a passenger during a driving simulation and record how many words they say during "dangerous" scenarios. All 20 participants are 18, but only half of them have a driver's license. What statistical test would we use to compare the number or words spoken for people with and without licenses. | | | | | Selected Answer: | Independent samples t test | Answers: | Dependent samples t test | | Independent samples t test | | | | | * Question 2 10 out of 10 points | | | We want to know how having kids affects happiness. We surveyed 100 couples that were expecting a child in the next three months. We went back and surveyed those same 100 couples one year later. What test would we use to compare their previous scores with their current scores? | | | | | Selected Answer: | Dependent samples | Answers: | Dependent samples | | Independent samples | | | | | * Question 3 10 out of 10 points | | | We are doing a dependent samples t-test and we're in the middle of step 4 of our null hypothesis test. We determined that D̅ =...
Words: 4970 - Pages: 20
...1 Carolyn Porterfield E-text 8.48 & 8.64 Exercise 8.48 (a display ad is a large block of multicolored illustrations, maps, and text). The data (in square millimeters) are shown below: 0 260 356 403 536 0 268 369 428 536 268 396 469 536 162 338 403 536 536 130 (a) Construct a 95 percent confidence interval for the true mean. x-bar = 346.5 s = 170.378 t-critical value for 95% CI with df=19 = 2.093 E = 2.093*170.378/sqrt (20) = 2.0931.96*38.0976=79.74 95% CI : (346.5-79.74,346.5+79.74) (b) Why might normality be an issue here? The Confidence Interval is a statement about the whole population. The random sample is probably not representative of the whole population. (c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence? n = [t*s/E] n = [2.093*170.378/10]^2 = 1271.64 n = 1272 (when rounded up) (d) If this is not a reasonable requirement, suggest one that is. 2 Increase E or decrease the confidence level; both will have the effect of lowering "n". Exercise 8.64 (a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop. We know N, the sample size, from the problem: N = 773 From the problem's givens, we can get p and q: p = 86/773 = 0.1113 q = 1 - p = 0.8887 From a z table, the value for the 90% interval is: 1.6449 Use the formula for the interval around a proportion: p - z*sqrt(pq/N) to p + z*sqrt(pq/N) 0.1113 -...
Words: 488 - Pages: 2
...Task 6 Statistics II February 4th 2013 A. Excel is used generate a random number using the RANDBETWEEN(1,4) method. Please see part A below. 1. To generate this sample from the whole population of job applicants, using Excel and the method RANDBETWEEN(1,4) the number 2 was generated. The RANDBETWEEN(1,4) method uses an algorithm to generate a random number. This method generates a large number using a seed that is based on the time noted in the computer time. This number is then reduced to the specified 1, 2, 3, or 4 using the mod 3 + 1. Using the random list that was provided, I was able to use systematic sampling to select every second data value and generated a sample. To do this, I started with the second number, in the random list provided, and then picked every second number to generate a sample that was 50 numbers large (shown above). B. 1,2,3 Using excel and the methods STDEV(), AVERAGE(), MODE(), and MEDIAN() so that the standard deviation, mean/average, mode and median are calculated. Examples of these are shown below for unsuccessful applicants the process is the same for successful applicant using the other data set. To find the mean using the AVERAGE method, Excel finds the sum of all the numbers selected and divides them by the number of data points, in this case 50. Example Mean: n=1n=50Xn=2184 Where Xn=Unsuccessful applicant Mean=218450=43.68 For the MODE() method, Excel looks over all the data that is selected and finds the value...
Words: 2290 - Pages: 10
...金融计量第三次作业(本科+硕士) 要求: A. 提交截止时间:2014 年 4 月 23 日; B. 将所有数据处理工作写入 dofile,名称为 W3_姓名_学号.do; C. 所有分析过程和结果写入 word 文档,按照《经济学(季刊)》的格式排版,名称为 W3_ 姓名_学号.docx。注:这是我评分的主要依据。 D. 将上述两份文档(无需压缩,以便于我直接打开查看)发送至:ef_lianyujun@163.com。 E. 邮件标题为:本科_W3_姓名_学号(本科生) ;硕士_W3_姓名_学号(硕士生) 。不按照 我的要求填写邮件标题可能导致作业漏登。 F. 请勿发送数据文档。 你的任务: 目的:研究中国民营企业高管薪酬的影响因素。 1. 请查阅国内外相关文献,对高管薪酬影响因素方面的文献做简要评述(500-1000 字) 。 评述的主要内容包括: 文献中主要从哪些角度研究了高管薪酬问题?(2) 影响高管薪 (1) 酬的主要因素有哪些?(3) 现有文献中有关高管薪酬决定因素的研究中,已经比较一致 的观点有哪些?尚存争议的观点有哪些? 2. 统计分析各个一级行业(SIC)在各个年度上的平均 CEO_compen,并作简要分析; 3. 在每个年度上,分别根据营业收入 (Sale)、负债率、ROA 和 Salegr 将样本分成两组: High 和 Low 组。 统计基于不同指标分组后两组公司的 CEO_compen 的平均值; 计 (1) (2) 算两组均值差异,并采用 t-test 检验均值差异是否显著。 (提示:t-test 可以使用 ttest 命 令,亦可使用外部命令 ttable2 或 ttestplus 等) 。表格形式如下: 分组指标 Sale Leverage ROA Salegr High Mean1 Mean1 Mean1 Mean1 Low Mean2 Mean2 Mean2 Mean2 difference Mean1-Mean2 Mean1-Mean2 Mean1-Mean2 Mean1-Mean2 t-value*** 4. 设定线性回归模型, 被解释变量为 CEO_compen (在第二次作业中已经定义了这个变量)。 (1) 解释变量可以根据你查阅的文献自行设定,说明你选择的变量反映了哪些方面的影 响(如经营业绩、资本结构、资产结构、行业特征、年度特征、公司治理、经理人个人 特征等)?(2) 对于选入的每个解释变量都要给出变量的定义、计算方法以及所参考的 主要文献。(3) 样本的筛选过程和离群值的处理方法需要详细说明。 提示:建议设定如下模型形式: CEO _ compenit a 1 X it s t it (1) 其中,CEO_compenit 表示公司 i 在第 t 年的高管薪酬,X 为一系列你认为可能影响高管 薪酬的解释变量, s 表示 S–1 个行业虚拟变量(基于上次作业中定义的一级行业分类变量 SIC 产生) t 表示 T–1 个年度虚拟变量吗, it 为干扰项。 , 5. 基本统计量的呈现。呈现模型中涉及的所有变量的基本统计量(Mean, SD, median, min, max, N) ,并将结果输出整理到 word 文档中,并作简要分析。提示:建议使用 tabstat 命令进行统计,进而配合使用 logout 命令输出到 Excel 中,稍作整理后贴入 Word。 1 6. 变量的相关系数矩阵。 使用 pwcorr 或 pwcorr_a...
Words: 316 - Pages: 2
...Bob’s Candies Bob loves making candy, especially varieties of caramel, including plain, chocolate dipped caramels and chocolate dipped caramels with pecans. Bob has received lots of compliments from his friends and neighbors, and several have encouraged him to start his own candy making business. After several days of research, Bob finds that the national average amount of money spent annually per person on this type of specialty candy is $75. Bob believes that the citizens in his area spend more than that per year. Knowing whether or not this is true could help Bob make a wise decision regarding his future business plans. Bob wants to use statistics to support his claim, and to help him obtain a small business loan. Bob also wants to find an estimate of the true amount of money local citizens do spend on this type of specialty candy. Bob randomly selects several people from his local phone book and asks the person that answers how much money they typically spend per year on candy like he will make. He obtains the following results (in dollars): 75, 74, 80, 68, 79, 85, 77, 82, 79, 67, 90, 72, 76, 75, 69, 85, 78, 79, 82, 66, 75, 85, 90, 76, 85, 67, 89, 82, 69, 79, 82, 80, 84, 79, 78, 81, 77, 84, 80, 76. Based upon these results, Bob is hoping his area has a good customer base for his new business. Bob also hopes the bank is impressed with his use of statistics and will grant him the loan he needs to start it! Questions: 1. What is an appropriate set of...
Words: 380 - Pages: 2
...Prepare answers to the following assignments from the e-text, Applied Statistics in Business and Economics, by Doane and Seward: Chapter 8 – Chapter Exercises - Like 8.48 and 8.64 but with different Numbers Chapter 8 Like 8.48 But with different numbers: A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured (a display ad is a large block of multicolored illustrations, maps, and text). The data (in square millimeters) are shown below: 0 520 712 806 1072 0 536 738 856 1072 536 792 938 1072 324 676 806 1072 1072 260 (a) Construct a 95 percent confidence interval for the true mean. Please show your formulas and all calculations. Must use the equation editor. Sample mean is: x=0+520+712+806+1072+0+536+738+856+1072+536+792+938+1072+324+676+806+1072+1072+26020=13,86020=693 Sample standard deviation is: xi | (xi-x)2 | 0 | 480,249 | 520 | 29,929 | 712 | 361 | 806 | 12,769 | 1072 | 143,641 | 0 | 480,249 | 536 | 24,649 | 738 | 2,025 | 856 | 26,569 | 1072 | 143,641 | 536 | 24,649 | 792 | 9,801 | 938 | 60,025 | 1072 | 143,641 | 324 | 136,161 | 676 | 289 | 806 | 12,769 | 1072 | 143,641 | 1072 | 143,641 | 260 | 187,489 | Total | 2,206,188 | 2,206,188(20-1)=340.76 Degrees of freedom are: v=n-1=20-1=19 Appendix D: t0.95=2.093 95% confidence interval is: x±tsn=693±2.093340.7620=693±159.48 ...
Words: 563 - Pages: 3
...A.BANKING OBJECTIVE: To differentiate the banks based on the type of ownership on the following variables a. Profits/(Total Assets-NPA) b. NPA/Total Assets c. Advances/Deposits d. (Total Assets-NPA)/Number of Branches REQUIRED DATA: Data is collected from www.moneycontrol.com and www.rbi.org on Profits, Total Assets, NPAs, Total Liabilities, Deposits, Advances, Bank’s Capital, number of branches, Ownership Variables (Public , Private( Foreign and Domestic) DESCRIPTION: Data is collected on Profits, Total Assets, NPAs, Total Liabilities, Deposits, Advances, Bank’s Capital, number of branches, Ownership Variables (Public , Private( Foreign and Domestic). The return of the stock for each month is calculated by the following formula: Rt =closing value of stock on date t-closing value of stock on date (t-1)closing value of stock on date (t-1) Where Rt - return of the stock. 1) Profits/(Total Assets-NPA) public and private bank Queone= Profits/(Total Assets-NPA) Group Statistics | | ownership | N | Mean | Std. Deviation | Std. Error Mean | | 1 | 18 | .050343 | .1305157 | .0307629 | | 2 | 10 | .147825 | .4362644 | .1379589 | Descriptives | | | | | | | | | | N | Mean | Std. Deviation | Std. Error | 95% Confidence Interval for Mean | Minimum | Maximum | | | | | | Lower Bound | Upper Bound | | | 1 | 18 | .050343 | .1305157 | .0307629 | -.014561 | .115247 | .0056 | .5627 | 2 | 10 | .147825 | .4362644 | .1379589...
Words: 1597 - Pages: 7
...(CIB pdf formfields Demoversion) BASIC ANALYSIS: A GUIDE FOR STUDENTS AND RESEARCHERS ASSOCIATE PROFESSOR DR ERNEST CYRIL DE RUN DR LO MAY CHIUN HERIYADI KUSNARYADI BASIC ANALYSIS: A GUIDE FOR STUDENTS AND RESEARCHERS 1 (CIB pdf formfields Demoversion) PREAMBLE This book was originally written as notes for my students of EBQ2053 Research Methodology at Universiti Malaysia Sarawak. Nevertheless, as we looked through it and with the various courses and seminars that we have given, we began to realize that what was being said was universal for all researchers, either those just starting out at 2nd year university of seasoned well published researchers. We all need to know the basics. Nevertheless, at the same time, even seasoned researchers tend to forget some methods that they do not always use. Therefore the idea for this book, as a handout for students yet at the same time a quick guide and reference for the seasoned researcher. Please note that we are using SPSS v15 and AMOS v4. May it be of help to all who strive to better themselves. This book is dedicated to or my darling wife, Doren, and my dearest son, Walter. Associate Professor Dr Ernest Cyril de Run 16 November 2007 BASIC ANALYSIS: A GUIDE FOR STUDENTS AND RESEARCHERS 2 (CIB pdf formfields Demoversion) 1. What is SPSS? SPSS refers to computer software named Statistical Program for Social Sciences and it comes in various versions and adds on. It is software and not a method of analysis....
Words: 11336 - Pages: 46
...Running head: ORGANIZATIONAL COMMITMENT 1 Organizational Commitment of Part-Time and Full-Time Employees Julia A. Teahen Baker College ORGANIZATIONAL COMMITMENT Abstract In recent years many educational institutions have increased their use of part-time adjuncts, especially with the introduction of distance learning courses. With this growing use questions about the efficacy of part-time and distance faculty have arisen. This paper tests whether organizational commitment, as described by Mowday (1979), differs between three groups of faculty: full-time, part-time, and part-time who work full-time for another organization. The 2 authors used a 15 item, seven-point scale instrument to measure commitment of a sample of 479 full and part-time faculty at two midwestern universities. Those that worked full-time for another organization are more committed to organizations for whom they work part-time than are those working only part-time. No significant difference in commitment was observed between full-time and part-time employees. ORGANIZATIONAL COMMITMENT Organizational Commitment of Part-Time and Full-Time Employees With the rapid growth of non-traditional educational institutions and enrollment, many educational institutions have added significant numbers of part-time adjuncts to teach courses. With this growth a number of questions and rhetoric have arisen about the quality of instruction from part-timers (Fulton, 2000; Leatherman, 1998; Rewick, 2001), "overuse...
Words: 3410 - Pages: 14