SPSS Guide by Jafri

SPSS GUIDE by Jafri Solved Examples of Different Statistical Techniques from Text Books β Σ λ α Σ σ γ χ π MUHAMMAD KAZIM JAFRI, 2009 m.kazimjafri9@gmail.com Department of Statistics University of Sindh Jamshoro 0|Page Contents Data Representation Topic # 1- Pie Chart Topic # 2: Dot Plot Topic # 3: Box Plot Topic # 4: Cumulative Frequency Polygon Topic # 5: Scatter Diagram Topic # 6: Scatter Plot Matrix Descriptive Statistics Topic # 7: Confidence Interval of Mean Topic # 8: Harmonic Mean Topic # 9: Trimmed Mean Topic # 10: Standard Scores Topic # 11: Standard Error of Mean Probability Distributions Topic # 12: Geometric Distribution Topic # 13: Exponential Distribution Topic # 14: Normal Distribution Topic # 15: Uniform Distribution Correlation Topic # 16: Correlation for Grouped data Topic # 17: Rank Correlation Topic # 18: Rank Correlation for Tied Ranks Topic # 19: Covariance Regression Analysis Topic # 20: Test of Linearity Topic # 21: Mean Prediction Topic # 22: Individual Prediction 1|Page Topic # 23: Confidence Interval on 𝛽1and 𝛽2 Topic # 24: Testing of Hypothesis about the Regression Coefficient β Topic # 25: ANOVA for Simple Linear Regression Topic # 26: ANOVA for Multiple Linear Regressions Topic # 27: Standard Error of Estimates Topic # 28: Coefficient of Determination Topic # 29: Coefficient of Multiple Determinations Topic # 30: Dummy Variable Topic # 31: Inverse Curve Design of Experiment Topic # 32: Latin – Square Design Topic # 33: The Graeco – Latin Square Design Topic # 34: Balance Incomplete Block Design Topic # 35: Youden Square Design Topic # 36: Completely Randomized Design Topic # 37: Randomized Complete Block Design Topic # 38: 𝟐𝟐 Factorial Experiment Topic # 39: 𝟐𝟑 Factorial Experiment Topic # 40: 𝟐𝟒 Factorial Experiment Topic # 41: Blocking in Factorial Design Topic # 42: Least Significant Difference (LSD) Topic # 43: Duncan’s Test Topic # 44: Tukey’s Test Topic # 45: Regression of 𝟐𝟑 Factorial Design Topic # 46: Dunnett’s Test Topic # 47: Test of Equality of Variance Topic # 48: Effect Size of One – Way (Eta-Square) Quality Control Topic # 49: Control Chart by the Cases in Units Topic # 50: Control Chart by the Sub – Groups Curvilinear Regression Topic # 51: Second Degree Curve Parabola Topic # 52: Third Degree Curve Parabola Topic # 53: Exponential Curve Topic # 54: Power Curve 2|Page Chi-Square Tests Topic # 55: Testing of Hypothesis about the P’s of Multinomial – Distribution Topic # 56: Goodness of Fit Test Topic # 57: Testing of Hypothesis about the Equality of Several Proportions Topic # 58: The Chi-Square Test as Test of Homogeneity Time Series Analysis Topic # 59: Fit the Second Degree Trend Curve Topic # 60: Growth Model Topic # 61: Autocorrelation Function (ACF) Topic # 62: Partial Autocorrelation Correlation Faction (PACF) Topic # 63: Time Series Graph Topic # 64: Exponential Smoothing Please Subscribe our Channel for Statistics and Economics Learning and Solutions https://www.youtube.com/channel/UCVOgn7b-ZGTUJUHWQ0I3miw/featured 3|Page Books’ Key Book # 1  Int to Stat-theory  Ex #  Pg #  Prb #  Author = Introduction to Statistical Theory (Part – I) = Example number = Page number = Problem number = Sher Muhammad Chaudhry Book # 2  Int to Stat-theory  Ex #  Pg #  Prb #  Author = Introduction to Statistical Theory (Part – II) = Example number = Page number = Problem number = Sher Muhammad Chaudhry Book # 3  Basic Eco  Sec #  Pg #  Author = Basic Econometrics = Section number = Page number = Damodar N. Gujrati Book # 4  Stat for Eco & Mngmnt  Ex #  Pg #  Prb #  Author = Statistics for Economics and Management = Example number = Page number = Problem number = Hamid A. Hakeem Book # 5  Design & experiment  Ex #  Pg #  Author = Design and analysis of experiment = Example number = Page number = Douglas C. Montgomery Book # 6  Int stat  Ex #  Sec #  Pg #  Author = Introductory Statistics = Example number = Section number = Page number = Neil A. Weiss 4|Page Book # 7  Busi Stat  Sec #  Pg #  Author = Business Statistics = Section number = Page number = Chris Robertson Book # 8  Int to stat  Ex #  Sec #  Pg #  Author = Introduction to Statistics = Example number = Section number = Page number = Ronald E. Walpole Book # 9  Txt Busi Stat  Ex #  Pg #  Author = Text Book of Business Statistics = Example number = Page number = A. K. Sharma Book # 10  Int linear Reg  Ex #  Pg #  Author = Introduction To Linear Regression Analysis = Example number = Page number = Douglas C. Montgomery Book # 11  Statistics  Ex #  Pg #  Author = Statistics = Example number = Page number = Mc Clave. Dietrich. Sincich Book # 12  Statistics bhvrl Scnc  Tbl #  Pg #  Author = Statistics for the Behavioral Science = Table number = Page number = Joan Welkowitz, Barry H. Cohen, Robert 5|Page Topic # 1: Pie Chart Represent the total expenditure and expenditures on various items of a family by a pie chart. Items Food Clothing House rent Fuel and Light Misc Expenditure 50 30 20 15 35 (Int to Stat-theory, P-I) Ex # 2.13, Pg # 36 From the menus choose: Analyze Descriptive Statistics Frequencies … Clothing Food House Rent Fuel & light Miscellaneous Topic # 2: Dot plots A farmer interested in estimating his yield of oats if he farms organically. He uses the method on a sample of 15 one – acre plots. The yields, in bushels, are depicted in following table. Construct a dot plot for the data. 6|Page 67 Oats Yields 65 55 57 58 61 61 61 64 62 62 60 62 60 67 (Int Stat) Ex # 2.13, pg # 84 From the menus choose: Graphs Legacy Dialogs Scatter/ Dot … (simple dot) Oat Yields 55 58 60 62 65 68 Yield (bushels) Topic # 3: Box plots The capital assets ratio presented in the following table. Construct the Box plot. 3.6 5.4 3.6 3.2 2.7 US UK Ja WG Ja 3.7 3.1 2.8 5.8 4.4 US WG WG SA US 5.8 3.5 4.0 6.9 3.2 Fr Ja Ja SA It 1.4 3.6 5.3 3.3 3.2 Fr US UK Ca Ja 1.1 3.8 3.0 3.5 7.0 Fr Ja US Nc UK 1.6 3.6 0.4 2.8 3.6 Fr Ja Ja WG Ja (Busi Stat) Sec # 3.4, pg # 47 From the menus choose: Graphs Legacy Dialogs Boxplots … (simple) 7|Page 14 25 9 6.0 11 4.0 2.0 16 21 28 0.0 Capital_assest Topic # 4: Cumulative Frequency Polygon Construct the cumulative frequency polygon by using the following data. Car Battery Lives 2.2 4.1 3.5 4.5 3.2 3.7 3 2.6 3.4 1.6 3.1 3.3 3.8 3.1 4.7 3.7 2.5 4.3 3.4 3.6 2.9 3.9 3.1 3.3 3.1 3.7 4.4 3.2 4.1 1.9 3.4 4.7 3.8 3.2 2.6 3.9 3 4.2 3.5 3.3 Make the frequency distribution at first of the ranges are, 1.5 – 1.9, 2.0 – 2.4, 2.5 – 2.9, 3.0 – 3.4, 3.5 – 3.9, 4.0 – 4.4 and 4.5 – 4.9. (Int to stat) sec # 3.2, Pg # 56 From the menus choose: Transform Recode into Different Variables … Graphs Legacy Dialogs Line … (simple) 8|Page Cumulative Frequency 40 30 20 10 0 1.5 - 1.9 2.0 - 2.4 2.5 - 2.9 3.0 - 3.4 3.5 - 3.9 4.0 - 4.4 4.5 - 4.9 new Topic # 5: Scatter Diagram Construct the Scatter Diagram by using the following data, in which the X represents the age of cars in the years, and Y represents the price of cars in the dollar ($). Car 1 2 3 4 5 6 7 8 9 10 11 Age (Years) 5 4 6 5 5 5 6 6 2 7 7 Price (dollar $) 85 103 70 82 89 98 66 95 169 70 48 (Int Stat) Sec # 14.2, pg # 827 From the menus choose: Graphs Legacy Dialogs Scatter/ Dot … (simple scatter) 9|Page 175 Price_Y 150 125 100 75 50 2 3 4 5 6 7 Age_X Topic # 6: Scatter Plot Matrix Draw a diagram of scatter plot matrix of 30 companies of from Fortune 500, November 1999. Data set is given in the following table: Company General motors Ford Waal mart Exxon General electric IBM Citigourp Phillip morris Boeing at & t bank of America Mobil Hewlet packart State frame ins cos Sears roebuck Ei du pont de numer Protect and gambol Tiaa-cref Revenue 1.8 15.3 3.2 3.6 9.2 7.8 7.6 9.3 2.0 11.2 1.2 3.6 6.3 3.0 2.5 11.4 10.2 2.3 Asset 1.1 9.3 9.0 6.9 2.6 7.3 .9 9.0 3.0 10.7 .8 4.0 8.8 1.2 2.8 11.3 12.2 .3 Equity 19.7 94.3 21.0 14.6 23.9 32.6 13.6 33.2 9.1 25.1 11.2 9.3 17.4 3.2 17.3 31.5 30.9 13.3 Earning 4.18 17.76 1.98 2.55 2.80 6.57 2.43 2.20 1.15 3.55 2.90 2.10 2.77 . 2.68 3.90 2.56 . Percentage 21.4 86.8 107.7 22.4 41.0 77.5 -6.8 22.8 -32.4 26.2 1.3 24.3 18.7 . -4.4 -9.7 15.9 . 10 | P a g e Merrill lynch Prudential Kmart American international chase Manhattan corp Texaco Bell atlantic Funnie mae Enron Compaq Morgan Stanley dean Dayton Putson 3.5 3.2 1.7 11.3 11.7 1.8 9.4 10.8 2.2 -8.8 10.5 3.0 .4 .4 4.0 1.9 1.0 2.0 5.4 .7 2.4 -11.9 1.0 6.0 12.4 5.4 9.5 13.9 15.9 4.9 22.7 22.1 10.0 -24.2 23.2 17.6 3.00 . 1.01 3.57 4.27 .99 1.86 3.23 2.02 -1.71 5.33 1.98 -7.5 . 33.1 33.6 32.5 .5 22.5 31.7 39.9 49.0 21.4 62.1 (Busi Stat) Sec # 3.7.2, pg # 60 From the menus choose: Graphs Legacy Dialogs Scatter/Dot … (matrix) revenue company percentage earning equity asset american internation at & t bank of america bell atilantic boeing chase manhattan corp citigourp compaq dayton putson ei du pont de numer enron exxon ford funnie mae general electric genral motors hewlet packart ibm kmart merrill lynch mobil morgan stanley dean phillip morris protect and gambol sears roebuck texaco wal mart revenue asset equity earning percentage 11 | P a g e Topic # 7: Confidence Interval for µ An Astronomer interesting in getting mean brightness of certain star got 6 reading on the stars during his experiment. Calculate 99% confidence interval for the true value of the population mean. Reading Brightness 1 30 2 32 3 29 4 31 5 33 6 30 (Stat for Eco & Mngmnt) Ex # 12.4, Pg # 308 From the menus choose: Analyze Descriptive Statistics Explore … Descriptives X Mean 99% Confidence Interval for Mean Statistic 30.83 28.41 Lower Bound Upper Bound Std. Error .601 33.26 5% Trimmed Mean 30.81 Median 30.50 Variance 2.167 Std. Deviation 1.472 Minimum 29 Maximum 33 Range 4 Interquartilae Range 3 Skewness Kurtosis .418 .845 -.859 1.741 Topic # 8: Harmonic Mean Compute the Harmonic Mean from the given data Salary level Hourly rate No-of-Employee 1 45 25 2 50 20 3 55 15 4 60 10 5 65 5 (Stat for Eco & Mngmnt) Prb # 50, Pg # 70 12 | P a g e From the menus choose: Data Weight cases … Analyze Compare Means Means … Report Harmonic Mean category A hourly rate 50.96 Total 50.96 Topic # 9: Trimmed Mean Calculate the trimmed mean for the given data 45, 32, 37, 46, 39, 36, 41, 48 and 36. (Int to Stat-theory, P-I) Ex # 4.12, Pg # 112 From the menus choose: Analyze Descriptive Statistics Explore … Descriptive X Statistic 40.00 Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Lower Bound Upper Bound 35.86 44.14 40.00 39.00 29.000 5.385 Minimum 32 Maximum 48 Range 16 Interquartile Range Skewness Kurtosis Std. Error 1.795 10 .191 .717 -1.161 1.400 Topic # 10: Standard Scores (z Scores) The distribution of the 20 base line heart rate scores expressed in beats per minute (bpm), compute the Z-Scores on base line distribution. 13 | P a g e Heart rate (bpm) 72 70 70 70 69 69 68 68 68 68 Male 1 2 3 4 5 6 7 8 9 10 Male 11 12 13 14 15 16 17 18 19 20 Heat rate (bpm) 67 67 67 67 67 67 66 66 66 64 (Statistics bhvral Scnc) Tbl # 7.1, Pg # 99 From the menus choose: Analyze Descriptive Statistics Descriptives … (save stnd values) Analyze Reports Case Summaries … Z score of Heart Rate 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Total Heart rate 72 70 70 70 69 69 68 68 68 68 67 67 67 67 67 67 66 66 66 64 72 Zscore(x) 2.30 1.21 1.21 1.21 .66 .66 .11 .11 .11 .11 -.44 -.44 -.44 -.44 -.44 -.44 -.99 -.99 -.99 -2.08 2.30 14 | P a g e Topic # 11: Standard of Mean A random sample of n = 6 has the elements 6, 10, 13, 14, 18 and 20 compute the standard error of the mean. (Int to Stat-theory, P-II) Ex # 15.1, Pg # 69 From the menus choose: Analyze Descriptive Statistics Descrivtives … Descriptive Statistics N Mean Statistic Y 6 Valid N (list wise) 6 Statistic 13.50 Std. Error 2.094 Topic # 12: Geometric Distribution If the probability that a person will believes a rumor about the retirement of certain politician is 0.25, what is the probability that. The 6 people to hear the rumor will be the first to believe it. (Int to Stat-theory, P-I) Ex # 8.26, Pg # 346 From the menus choose: Transform Compute Variable … (PDF) Analyze Reports Case Summaries … Geometric Distribution 1 x 6 probability .059 Total 6 .059 Topic # 13: Exponential Distribution The duration of the long distance telephone calls is found to be exponentially distributed with a mean of 3 minutes. The mean of distribution is 3, so that λ = 1 / 3.What is the probability that a call will last, 1. More than 3 minutes. 2. More than 5 minutes? (Int to Stat-theory, P-I) Ex # 9.1, Pg # 365 15 | P a g e Note: Here we use the cumulative distribution function (CDF) because, the probability is required more than 3 minutes or 5 minutes. CDF calculate the probability less than the given value therefore we subtract from total probability “1”. From the menus choose: Transform Compute Variable … (CDF) Analyze Reports Case Summaries … Exponential distribution 1 x 3 Probability .3679 2 5 .1889 Total 3 .3679 Topic # 14: Normal Distribution Give a normal distribution with µ = 300 and σ = 50, find the probability that X assumes a value greater than 362. (Int to stat) Ex # 2, Pg # 189 Note: Here we use the cumulative distribution function (CDF) because, the probability is required greater than 362. CDF calculate the probability less than the given value therefore we subtract from total probability “1”. From the menus choose: Transform Compute Variable … (CDF) Analyze Reports Case Summaries … Normal Distribution 1 x 362 probability .1075 Total 362 .1075 16 | P a g e Topic # 15: Uniform Distribution An unprincipled used car dealer sells a car to an unsuspected buyer, even thought the dealer knows that the car will have a major breakdown within the next 6 months. Assume that x is a uniform random variable with values between 0 and 6 months. We must find the probability that between the points x = 0 and x = 1.5. P (0 < x < 1.5) =? (Statistics) Ex # 5.1, Pg # 196 From the menus choose: Transform Compute Variable … (CDF) Analyze Reports Case Summaries … UNIFORM DISTRIBUTION 1 X 1.5 probability .25 Total 1.5 .25 Topic # 16: Correlation for Grouped Data Compute correlation coefficient from the following correlation table for weights and heights of women students. Heights in inches 57 60 63 66 69 72 90 --8 3 1 ----- 110 --21 50 24 1 --- Weights in pounds 130 150 ----8 1 57 12 54 19 8 5 2 --- 170 1 ----3 3 --- 190 ----2 ----1 (Int to Stat-theory, P-I) Prb # 10.26, Pg # 455 From the menus choose: Data Weight cases … Analyze Descriptive Statistics Crosstabs … 17 | P a g e Symmetric Measures Asymp. Std. Error(a) Value .388 Interval by Interval Pearson's R Ordinal by Ordinal Spearman Correlation .395 N of Valid Cases Approx. T(b) Approx. Sig. .065 7.068 .000(c) .053 7.213 .000(c) 284 Topic # 17: Rank Correlation Find the coefficient of rank correlation from the following ranking of six values between two data sets? A B 7.4 8.5 9.0 6.1 11.0 2.4 2.5 6.7 4.6 12.6 6.5 3.3 (Int to Stat-theory, P-I) Prb # 10.30, Pg # 455 From the menus choose: Analyze Correlate Bivariate … Correlations A Spearman's rho A Correlation Coefficient Sig. (2-tailed) N B Correlation Coefficient Sig. (2-tailed) N 1.000 B -.600 . .208 6 -.600 1.000 6 .208 . 6 6 Topic # 18: Rank Correlation for Tied Ranks Two members of selection committee rank eight persons according to their suitably for promotion as follows. Persons Member – 1 Member – 2 A 1 2 B 2.5 4 C 2.5 1 D 4 3 E 5 6 F 6 6 G 7 6 H 8 8 Calculate the coefficient rank correlation? (Int to Stat-theory, P-I) Ex # 10.11, Pg # 446 18 | P a g e From the menus choose: Analyze Correlate Bivariate … Correlations member1 Spearman's rho member1 Correlation Coefficient 1.000 member2 .896 Sig. (2-tailed) . .003 8 Correlation Coefficient 8 .896 1.000 Sig. (2-tailed) .003 . 8 8 N member2 N Topic # 19: Covariance Find out the covariance between the following variables X and Y are shown in the table, Country Japan Italy UK W. Germany USA Spain Australia New Zealand X 0.28 0.33 0.41 0.41 0.43 0.44 0.49 0.71 Y 3.7 2.9 2.2 2.4 2.1 1.6 1.8 0.3 (Busi Stat) Sec # 6.4.1, pg # 128 From the menus choose: Analyze Scale Reliability Analysis … (covariance) Inter-Item Covariance Matrix X x y .016 -.122 y -.122 .982 19 | P a g e Topic # 20: Test Of linearity Of Regression The following data show the heights (X) and weight (Y) of twelve men. We selected the height in advance and then observed the weights of a random group of men having the selected heights. X (height) Y (weight) 60 60 60 62 62 62 62 64 64 70 70 70 110 135 120 120 140 130 135 150 145 170 185 160 Test the hypothesis at the 0.05 level of significance, that the regression is linear. (Int to Stat-theory, P-II) Ex # 21.7, Pg # 366 From the menus choose: Analyze Compare Means Means … ANOVA Table Y* X Between Groups Sum of Squares 4402.08 3 4341.81 2 (Combined) Linearity Deviation from Linearity Within Groups Total Mean Square df F Sig. 3 1467.361 13.578 .002 1 4341.812 40.175 .000 60.271 2 30.136 .279 .764 864.583 5266.66 7 8 108.073 11 Topic # 21: Mean Prediction On the basis of the sample data of 3.2 we obtained the following sample regression 𝑌𝑖 = 24.4545 + 0.5091𝑋𝑖 . Assume that 𝑋0 = 100 and we want to predict 95 % confidence interval for Mean prediction and Individual prediction. X ($) Y ($) 80 70 100 65 120 90 140 95 160 110 180 115 200 120 220 140 240 155 260 150 (Basic Eco) Sec # 5.10, Pg # 137 From the menus choose: Analyze Regression Linear … Analyze Reports Case Summaries … 20 | P a g e Mean prediction 2 80 100 70 65 95% L CI for Y mean 56.38144 67.89991 3 120 90 79.26830 91.82261 4 140 95 90.38565 101.06890 5 160 110 101.10304 110.71514 6 180 115 111.28486 120.89696 7 200 120 120.93110 131.61435 8 220 140 130.17739 142.73170 9 240 155 139.17264 154.10009 10 260 150 148.01781 165.61856 80 70 56.38144 73.98219 X 1 Y Total 95% U CI for Y mean 73.98219 82.82736 Topic # 22: Individual Prediction On the basis of the sample data of 3.2 we obtained the following sample regression 𝑌𝑖 = 24.4545 + 0.5091𝑋𝑖 . Assume that 𝑋0 = 100 and we want to predict 95 % confidence interval for Mean prediction and Individual prediction. X ($) Y ($) 80 70 100 65 120 90 140 95 160 110 180 115 200 120 220 140 240 155 260 150 (Basic Eco) Sec # 5.10, Pg # 137 From the menus choose: Analyze Regression Linear … Analyze Reports Case Summaries … Individual Prediction X 1 2 3 4 5 6 7 8 9 10 Total N Y 80 100 120 140 160 180 200 220 240 260 10 70 65 90 95 110 115 120 140 155 150 10 95% L CI for Y individual 47.81420 58.63358 69.30999 79.83009 90.18377 100.36559 110.37554 120.21909 129.90631 139.45057 10 95% U CI for Y individual 82.54943 92.09369 101.78091 111.62446 121.63441 131.81623 142.16991 152.69001 163.36642 174.18580 10 21 | P a g e Topic # 23: Confidence Intervals for Regression Coefficients β1 and β2 Construct the 95% confidence intervals on the regression coefficients β1 and β2 for the given data. X ($) Y ($) 80 70 100 65 120 90 140 95 160 110 180 115 200 120 220 140 240 155 260 150 (Basic Eco) Sec # 5.3, Pg # 117 From the menus choose: Analyze Regression Linear … Coefficients 95% Confidence Interval for B Model 1 Lower Bound 9.664 .427 (Constant) X Upper Bound 39.245 .592 Topic # 24: Testing of Hypothesis for Regression Coefficient β Estimate the regression line from the following data of height (X) and weight (Y) of 12 persons. Heights (X) 60 62 64 70 Weights (Y) 110,135,120 120,140,130,135 150,145 170,185,160 Test the hypothesis that the population regression coefficient β = 0. Use 0.05 level of significance. (Int to Stat-theory, P-II) Ex # 21.5, Pg # 359 From the menus choose: Analyze Regression Linear … Coefficients Unstandardized Coefficients Model 1 (Constant) X B -179.359 5.029 Std. Error 46.936 .734 Standardized Coefficients Beta .908 t B -3.821 6.852 Sig. Std. Error .003 .000 22 | P a g e Topic # 25: ANOVA for Simple Linear Regression and Test of Ho: β = 0 Test the hypothesis that β = 0 at the 0.05 level of significance for the given data by setting the results in an ANOVA table. Heights (X) 60 62 64 70 Weights (Y) 110,135,120 120,140,130,135 150,145 170,185,160 (Int to Stat-theory, P-II) Ex # 21.5, Pg # 359 From the menus choose: Analyze Regression Linear … ANOVA Model 1 Sum of Squares df Mean Square Regression 4341.812 1 4341.812 Residual 924.854 5266.667 10 11 92.485 Total F 46.946 Sig. .000(a) Conclusion: The value of the F statistic is 46.946 that fall in critical region, and the probability of significance (Sig) is 0.00, therefore we reject our null hypothesis and conclude that regression coefficient for simple linear regression is not zero, it has some specific slope. Topic # 26: ANOVA for Multiple Regression and Testing Hypothesis about the β Parameter Consider the following set of data: Y X1 X2 4 1 2 6 1 4 5 2 3 8 3 4 12 4 8 13 5 8 19 7 11 Obtain the ANOVA table and test the hypothesis that there is no association between both regressor and the dependent variables. Use the 0.01 level of significance. (Int to Stat-theory, P-II) Ex # 21.16, Pg # 386 23 | P a g e From the menus choose: Analyze Regression Linear … ANOVA Model 1 Sum of Squares 172.053 Regression Residual df 2 1.661 173.714 Total Mean Square 86.027 4 6 F 207.154 Sig. .000(a) .415 Conclusion: The value of the F statistic is 207.154 that fall in critical region, and the probability of significance (Sig) is 0.00, therefore we reject our null hypothesis and conclude that regression coefficient for multiple linear regression is not zero, it has some specific slope. Topic # 27: Standard Error of Estimate Using the following data, find the standard error of estimate 𝑆𝑦.𝑥 ? X Y 5 16 6 19 8 23 10 28 12 36 13 41 15 44 16 45 17 50 (Int to Stat-theory, P-I) Ex # 10.3, Pg # 431 From the menus choose: Analyze Regression Linear … Model Summary Model 1 R .994(a) R Square .987 Adjusted R Square .985 Std. Error of the Estimate 1.507 Topic # 28: Coefficient of Determination Using the following data, find the Coefficient of determination? X Y 5 85 4 103 6 70 5 82 5 89 5 98 6 66 6 95 2 169 7 70 7 48 (Int Stat) Ex # 14.7, pg # 845 24 | P a g e From the menus choose: Analyze Regression Linear … Model Summary Model 1 R .924(a) R Square .853 Adjusted R Square .837 Std. Error of the Estimate 12.577 Topic # 29: Coefficient of Multiple Determinations A statistician wants to predict the incomes of restaurant, using two independent variables the number of restaurant employees and restaurant floor area. He calculated the following data: Income (000) Y 30 22 16 7 14 Floor area (000 sq. ft) Number of employees 10 5 10 3 2 15 8 12 7 10 𝑋1 𝑋2 Compute the standard error of estimate, coefficient of multiple determinations? (Int to Stat-theory, P-I) Ex # 11.3, Pg # 463 From the menus choose: Analyze Regression Linear … Model Summary Model 1 R .738(a) R Square .545 Adjusted R Square .089 Std. Error of the Estimate 8.276 Topic # 30: Dummy Variable Twenty observations of tool life and lathe speed are presented in table. Where the hours are a dependent variable, use the dummy variable Tool type and estimate the regression equation. 25 | P a g e Speed_arpm (𝑋1 ) 610 950 720 840 980 530 680 540 890 730 670 770 880 1000 760 590 910 650 810 500 Hours (Y) 18.73 14.52 17.43 14.54 13.44 24.39 13.34 22.71 12.68 19.32 30.16 27.09 25.40 26.05 33.49 35.62 26.07 36.78 34.95 43.67 Where the Tool Type (𝑋2 ) A A A A A A A A A A B B B B B B B B B B 0 𝑖𝑓 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑟𝑜𝑚 𝑡𝑜𝑜𝑙 𝑡𝑦𝑝𝑒 𝐴 𝑋2 = { 1 𝑖𝑓 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑟𝑜𝑚 𝑡𝑜𝑜𝑙 𝑡𝑦𝑝𝑒 𝐵 (Int linear Reg) Ex # 8.1, Pg # 267 From the menus choose: Analyze Regression Linear … Note: before analyze the data, first we values labeled A as 0 and B as 1. Coefficients Unstandardized Coefficients Model 1 (Constant) x tool B 36.986 Std. Error Standardized Coefficients t Sig. Beta B Std. Error 10.536 .000 -.027 3.510 .005 -.451 -5.887 .000 15.004 1.360 .845 11.035 .000 26 | P a g e Topic # 31: Inverse Curve Fit the inverse curve on the following data, Year 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962 1963 1964 1965 1966 Unemployment 1.4 1.1 1.5 1.5 1.2 1.0 1.1 1.3 1.8 1.9 1.5 1.4 1.8 2.1 1.5 1.3 1.4 Wage rate 1.8 8.5 8.4 4.5 4.3 6.9 8.0 5.0 3.6 2.6 2.6 4.2 3.6 3.7 4.8 4.3 4.6 (Basic Eco) Sec # 6.7, Pg # 177 From the menus choose: Analyze Regression Curve Estimation … Model Summary and Parameter Estimates Model Summary Equation Inverse R Square .385 F 9.385 df1 Parameter Estimates df2 1 Sig. 15 .008 Constant -1.428 b1 8.724 27 | P a g e Topic # 32: Latin Square Design Five fertilizers A, B, C, D and E were tested by the arranging plants in a Latin Square design in a field. The rows and columns in the table are rows and columns in the field. The yields per plot are as shown COLUMNS 1 2 3 4 5 1 B 4.9 D 6.4 E 3.3 A 9.5 C 11.8 2 C 9.3 A 4.0 B 6.2 E 5.1 D 5.4 ROWS3 D 7.6 C 15.4 A 6.5 B 6.0 E 4.6 4 E 5.3 B 7.6 C 13.2 D 8.6 A 4.9 5 A 9.3 E 6.3 D 11.8 C 15.9 B 7.6 Analyze the data for evidence at 5% level that the mean yields are not equal for the five fertilizers. (Int to Stat-theory, P-II) Ex # 23.6, Pg # 446 From the menus choose: Analyze General Linear Model Univariate … Tests of Between-Subjects Effects Source Corrected Model Intercept Rows Columns Type III Sum of Squares df Mean Square 21.441 9.175 .000 1544.490 1 1544.490 660.886 .000 46.668 4 11.667 4.992 .013 1.500 21.032 .263 14.020 4 4 3.505 49.152 Error 28.044 12 2.337 Total 1829.830 285.340 25 24 Corrected Total Sig. 12 196.608 treatments F 257.296(a) .000 Conclusion: Since the computed value of F = 21. 03 falls in the critical region, we therefore reject our null hypothesis. Hence we conclude that the data provide sufficient evidence to indicate at the 5% significance level that the mean yields for the fertilizers are not equal. 28 | P a g e Multiple Comparisons LSD Mean Difference (I-J) .380 -6.280(*) Std. Error .9669 .9669 Sig. .701 .000 Lower bound -1.727 -8.387 Upper bound 2.487 -4.173 D -1.120 .9669 .269 -3.227 .987 E 1.920 .9669 .070 -.187 4.027 A -.380 .9669 .701 -2.487 1.727 C D E -6.660(*) -1.500 1.540 .9669 .9669 .9669 .000 .147 .137 -8.767 -3.607 -.567 -4.553 .607 3.647 A 6.280(*) .9669 .000 4.173 8.387 B 6.660(*) .9669 .000 4.553 8.767 D 5.160(*) .9669 .000 3.053 7.267 E A B 8.200(*) 1.120 1.500 .9669 .9669 .9669 .000 .269 .147 6.093 -.987 -.607 10.307 3.227 3.607 C -5.160(*) .9669 .000 -7.267 -3.053 E 3.040(*) .9669 .008 .933 5.147 A -1.920 .9669 .070 -4.027 .187 -1.540 .9669 -8.200(*) .9669 -3.040(*) .9669 * The mean difference is significant at the .05 level. .137 .000 .008 -3.647 -10.307 -5.147 .567 -6.093 -.933 (I) treatments A B C D E (J) treatments B C 95% Confidence Interval B C D Topic # 33: The Graeco–Latin Square Design Suppose that in the dynamite formulation experiment of example an additional factor, test assemblies, could be of importance. Let there be five test assemblies denoted by the Greek letters α (alpha), β (beta), γ (gamma), δ (delta) and ε (epsilon). The resulting 5×5 Graeco – Latin Square design is shown in following table. Operators 1 2 3 4 5 1 A α -1 B γ -5 C ε -6 D β -1 E δ -1 2 B β -8 C δ -1 Dα 5 Eγ 2 A ε 11 Raw Material 3 C γ -7 D ε 13 Eβ 1 Aδ 2 B α -4 4 Dδ 1 Eα 6 Aγ 1 B ε -2 C β -3 5 E ε -3 Aβ 5 B δ -5 Cα 4 Dγ 6 (Design & experiment) Ex # 4.4, Pg # 152 29 | P a g e From the menus choose: Analyze General Linear Model Univariate … Tests of Between-Subjects Effects Source Corrected Model Intercept Type III Sum of Squares df Mean Square 610.000(a) 16 38.125 F Sig. 4.621 .017 4.000 1 4.000 .485 .506 Raw Material 68.000 4 17.000 2.061 .178 Operators Formulation 150.000 4 37.500 4 82.500 4.545 10.000 .033 330.000 62.000 4 15.500 1.879 .208 8.250 Greek Error 66.000 8 Total 680.000 676.000 25 24 Corrected Total .003 Conclusion: Since the computed value of F = 10 falls in the critical region, we therefore reject our null hypothesis. Hence we conclude that the data provide sufficient evidence to indicate at the 5% significance level that the mean yields for the Formulation are not equal. Multiple Comparisons LSD (I) Formulation A (J) Formulation B C D B C Mean Difference (I-J) 8.40(*) 6.20(*) Std. Error 1.817 1.817 Sig. .002 .009 Lower bound 4.21 2.01 Upper bound 12.59 10.39 -5.39 2.99 95% Confidence Interval -1.20 1.817 .527 E 2.60 1.817 .190 -1.59 6.79 A -8.40(*) 1.817 .002 -12.59 -4.21 C D E -2.20 -9.60(*) -5.80(*) 1.817 1.817 1.817 .260 .001 .013 -6.39 -13.79 -9.99 1.99 -5.41 -1.61 A -10.39 -2.01 -6.20(*) 1.817 .009 B 2.20 1.817 .260 -1.99 6.39 D -7.40(*) 1.817 .004 -11.59 -3.21 E A B -3.60 1.20 9.60(*) 1.817 1.817 1.817 .083 .527 .001 -7.79 -2.99 5.41 .59 5.39 13.79 C 7.40(*) 1.817 .004 3.21 11.59 E 3.80 1.817 .070 -.39 7.99 A -2.60 1.817 .190 -6.79 1.59 5.80(*) C 3.60 D -3.80 * The mean difference is significant at the .05 level. 1.817 1.817 1.817 .013 .083 .070 1.61 -.59 -7.99 9.99 7.79 .39 D E B 30 | P a g e Topic # 34: Balance Incomplete Block Design Consider the data in the given table for the Catalyst experiment. This is a balanced incomplete block design with a = 4, b = 4, k = 3, r = 3, λ = 2 and N = 12. Construct the ANOVA table. Block (Raw Material) 2 3 Treatment (Catalyst) 1 1 73 74 --- 71 2 --- 75 67 72 3 73 75 68 --- 4 75 --- 72 75 4 (Design & experiment) Ex # 4.5, Pg # 156 From the menus choose: Analyze General Linear Model Univariate … Tests of Between-Subjects Effects Source Corrected Model Intercept Type III Sum of Squares df Mean Square F Sig. 77.750(a) 6 12.958 19.936 .002 63075.000 1 63075.000 97038.462 .000 treatments 22.750 3 7.583 11.667 .011 Blocks 33.889 .001 66.083 3 22.028 Error 3.250 5 .650 Total 63156.000 12 81.000 11 Corrected Total Conclusion: The value of the F statistic is 11.667 that fall in critical region, and the probability of significance (Sig) is 0.01, therefore we reject our null hypothesis and conclude that the all treatment’s mean are not equal. 31 | P a g e Multiple Comparisons LSD Mean Difference (I-J) 1.33 .67 Std. Error .658 .658 Sig. .099 .358 Lower bound -.36 -1.03 Upper bound 3.03 2.36 4 -1.33 .658 .099 -3.03 .36 1 -1.33 .658 .099 -3.03 .36 3 4 1 -.67 -2.67(*) -.67 .658 .658 .658 .358 .010 .358 -2.36 -4.36 -2.36 1.03 -.97 1.03 2 .67 .658 .358 -1.03 2.36 4 -2.00(*) .658 .029 -3.69 -.31 1.33 .658 2 2.67(*) .658 3 2.00(*) .658 * The mean difference is significant at the .05 level. .099 .010 .029 -.36 .97 .31 3.03 4.36 3.69 (I) treatments 1 2 3 4 (J) treatments 2 3 95% Confidence Interval 1 Topic # 35: Youden Squares Design An industrial engineering is studying the effect of five illumination levels on the occurrence of defects in an assembly operation. Because time may be factor in the experiment, she has decided to run the experiment in five blocks, where each block is a day of the week. However, the department in which the experiment is conducted has four work stations, and these stations represent a potential source of variability. The engineer decided to run a Youden Square with five rows (block_days), four columns (work_station), and five treatments (illumination_level). Work Station 2 3 Day (Block) 1 1 A=3 B=1 C = -2 D=0 2 B=0 C=0 D = -1 E=7 3 C = -1 D=0 E=5 A=3 4 D = -1 E=6 A=4 B=0 5 E=5 A=2 B=1 C = -1 4 (Design & experiment) Ex # 6.2, Pg # 181 From the menus choose: Analyze General Linear Model Univariate … 32 | P a g e Tests of Between-Subjects Effects Source Corrected Model Intercept Type III Sum of Squares Illumination_level Mean Square F Sig. 128.417(a) 11 11.674 14.295 .000 48.050 1 48.050 58.837 .000 .867 4 .217 .265 .892 1.350 3 .450 .551 .662 120.367 4 30.092 36.847 .000 .817 Block_days Work_station df Error 6.533 8 Total 183.000 20 Corrected Total 134.950 19 Conclusion: The value of the F statistic is 36.847 that fall in critical region, and the probability of significance (Sig) is 0.00, therefore we reject our null hypothesis and conclude that the all illumination levels are not equal. Multiple Comparisons LSD (I) Illumination_level A B C D E 95% Confidence Interval Lower Upper bound bound 1.03 3.97 2.53 5.47 Mean Difference (I-J) 2.50(*) 4.00(*) Std. Error .639 .639 Sig. .004 .000 D 3.50(*) .639 .001 2.03 4.97 E -2.75(*) .639 .003 -4.22 -1.28 A -2.50(*) .639 .004 -3.97 -1.03 C D E 1.50(*) 1.00 -5.25(*) .639 .639 .639 .047 .156 .000 .03 -.47 -6.72 2.97 2.47 -3.78 A -4.00(*) .639 .000 -5.47 -2.53 B -1.50(*) .639 .047 -2.97 -.03 D -.50 .639 .456 -1.97 .97 E A B -6.75(*) -3.50(*) -1.00 .639 .639 .639 .000 .001 .156 -8.22 -4.97 -2.47 -5.28 -2.03 .47 C .50 .639 .456 -.97 1.97 E -6.25(*) .639 .000 -7.72 -4.78 A 2.75(*) .639 .003 1.28 4.22 .639 .639 .639 .000 .000 .000 3.78 5.28 4.78 6.72 8.22 7.72 (J) Illumination_level B C B 5.25(*) C 6.75(*) D 6.25(*) * The mean difference is significant at the .05 level. 33 | P a g e Topic # 36: The Complete Randomized Design. An experiment was conducted to compare the yields of varieties of potatoes. Each variety was assigned at random to equal size plot, four times. The yield were as follows, Variety A 23 26 20 17 B 18 28 17 21 C 16 25 12 14 Test the hypothesis that the three varieties of potatoes are not different in the yield capabilities. (Int to Stat-theory, P-II) Ex # 23.1, Pg # 424 From the menus choose: Analyze Compare Means One – Way ANOVA … ANOVA Sum of Squares Between Groups df Mean Square 54.500 2 27.250 Within Groups 217.750 9 24.194 Total 272.250 11 F 1.126 Sig. .366 Conclusion: Since the computed value of F = 1.13 does not fall in the critical region, so we accept our null hypothesis and may conclude that there is no difference among the yielding capabilities of the three varieties of potatoes. Topic # 37: The Randomized Complete Block Design. Four varieties of wheat were tried in a randomized complete block design in four replications. Yields in KG per plot are shown in the table given below. Test of hypothesis that there are no differences in the means of four varieties. Replicates I II III IV V1 2 2 4 1 Varieties V2 V3 5 4 3 3 6 6 4 2 V4 1 1 2 3 (Int to Stat-theory, P-II) Ex # 23.2, Pg # 429 34 | P a g e From the menus choose: Analyze General linear Model Univariate… Tests of Between-Subjects Effects Source Corrected Model Type III Sum of Squares df Mean Square F Sig. 31.875(a) 6 5.313 5.276 .014 150.063 1 150.063 .000 Replicates 12.188 3 4.063 149.028 4.034 Varieties 19.687 3 6.562 6.517 .012 1.007 Intercept Error 9.063 9 Total 191.000 16 40.938 15 Corrected Total .045 Conclusion: Since the computed value of F for the replicates and for the Varieties both falls in the critical region, we therefore reject our null hypotheses and conclude that the means of the Replications and means of the four varieties of wheat are significantly different. Furthermore, since F-statistic indicates rejection of the null hypothesis, we therefore apply the LSD on varieties to find out which means differ significantly. Multiple Comparisons LSD Mean Difference (I-J) -2.25(*) -1.50 Std. Error .710 .710 Sig. .011 .064 Lower bound -3.86 -3.11 Upper bound -.64 .11 4 .50 .710 .499 -1.11 2.11 1 2.25(*) .710 .011 .64 3.86 3 .75 2.75(*) 1.50 .710 .710 .710 .318 .004 .064 -.86 1.14 -.11 2.36 4.36 3.11 2 -.75 .710 .318 -2.36 .86 4 2.00(*) .710 .020 .39 3.61 -.50 .710 -2.75(*) .710 -2.00(*) .710 * The mean difference is significant at the .05 level. .499 .004 .020 -2.11 -4.36 -3.61 1.11 -1.14 -.39 (I) varieties 1 2 3 4 (J) varieties 2 3 4 1 1 2 3 95% Confidence Interval 35 | P a g e Topic # 38: 𝟐𝟐 Factorial Experiment The data of the following table are from a 2 × 2 factorial experiment. Partition the treatment sum of squares into main effects and interaction component. Treatment combinations A0b0 6 14 8 9 7 A0b1 12 14 13 11 13 a1b0 26 17 21 30 27 a1b1 21 16 20 17 21 (Int to Stat-theory, P-II) Prb # 23.37, Pg # 474 From the menus choose: Analyze General Linear Model Univariate … (full factorial) Tests of Between-Subjects Effects Source Corrected Model Intercept A B Type III Sum of Squares df Mean Square Sig. 3 232.583 21.535 .000 5216.450 1 5216.450 .000 594.050 1 594.050 483.005 55.005 2.450 1 2.450 .227 .640 101.250 10.800 9.375 .007 A*B 101.250 1 Error 172.800 16 Total 6087.000 870.550 20 19 Corrected Total F 697.750(a) .000 Conclusion: Since the computed values of F for the treatment combinations is 21.535 regarding treatment combinations falls in critical region., so we reject null hypothesis, that there is no significant difference between combinations. As null hypothesis is rejected, we therefore test the significance of differences between treatment A and treatment B, etc.  We reject hypothesis about treatment A’ effects are equal.  We accept hypothesis about treatment B’ effects are equal.  We reject hypothesis about there in no interaction between treatment A and B. 36 | P a g e Topic # 39: 𝟐𝟑 Factorial Experiment The following data were obtained from a 𝟐𝟑 factorial experiment repeated 3 times Replication 1 2 3 (1) A Treatment combinations b ab C ac bc 12 19 10 15 20 16 24 16 17 23 17 27 17 25 21 16 19 19 24 23 29 abc 28 25 20 (Int to Stat-theory, P-II) Prb # 23.41, Pg # 476 From the menus choose: Analyze General Linear Model Univariate … Tests of Between-Subjects Effects Source Corrected Model Intercept A Type III Sum of Squares 323.167(a) df 7 9680.167 1 Mean Square 46.167 F 3.095 Sig. .029 648.950 .000 2.667 1 9680.167 2.667 .179 .678 B 170.667 1 170.667 11.441 .004 C 104.167 1 104.167 6.983 .018 A*B 1.500 1 1.500 .101 .755 A*C 42.667 1 42.667 2.860 .110 B*C .000 1 .000 .000 1.000 1.500 1 1.500 .101 .755 Error 238.667 16 14.917 Total 10242.000 24 561.833 23 A*B*C Corrected Total Topic # 40: 24 Factorial Experiment A chemical product is produced in a pressure vessel. A factorial experiment is carried out in the plot plant to study the factor thought to influence the filtration rate of this product. The four factors are temperature (A), pressure (B), concentration of formaldehyde (C), and stirring rate (D). Perform 24 factorial experiment for following data. 37 | P a g e Run Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Run Label (1) A B ab C ac bc abc D ad bd abd cd acd bcd abcd Filtration Rate 45 71 48 65 68 60 80 65 43 100 45 104 75 86 70 96 (Design & experiment) Ex # 6.2, Pg # 246 From the menus choose: Analyze General Linear Model Univariate … Tests of Between-Subjects Effects Source Corrected Model Intercept A B C D A*B A*C B*C A*B*C A*D B*D A*B*D C*D A*C*D B*C*D A*B*C*D Error Total Corrected Total Type III Sum of Squares 5730.937(a) 78540.063 1870.563 39.063 390.063 855.563 .063 1314.063 22.563 14.063 1105.563 .563 68.063 5.063 10.563 27.563 7.563 .000 84271.000 5730.937 df Mean Square 15 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 16 15 382.062 78540.063 1870.563 39.063 390.063 855.563 .063 1314.063 22.563 14.063 1105.563 .563 68.063 5.063 10.563 27.563 7.563 . 38 | P a g e Topic # 41: Blocking in Factorial Design An engineer is studying methods for improving the ability to detect targets on a radar scope. Two factors she considers to be important are the amount of background noise, or “Ground Clutter” on the scope and “The Type of Filter” placed over the screen. Operator(blocks) Filter Types Ground Clutter Low Medium High 1 2 3 4 1 2 1 2 1 2 1 2 90 102 114 86 87 93 96 106 112 84 90 91 100 105 108 92 97 95 92 96 98 81 80 83 Perform the analysis and make ANOVA table. (Design & experiment) Ex # 5.6, Pg # 207 From the menus choose: Analyze General Linear Model Univariate … (interaction) Tests of Between-Subjects Effects Source Corrected Model Intercept Ground clutter Blocks filter types Ground clutter * filter types Error Total Corrected Total Type III Sum of Squares 1881.500(a) df 8 Mean Square 235.188 F 21.209 Sig. .000 216220.167 1 216220.167 2 167.792 19498.813 15.132 .000 335.583 .000 402.167 3 134.056 1 1066.667 12.089 96.192 .000 1066.667 77.083 2 38.542 3.476 .058 11.089 166.333 15 218268.000 24 2047.833 23 .000 Conclusion: Since the computed value of F for the Ground clutter and for the Filter types both falls in the critical region, we therefore reject our null hypotheses and conclude that the means of the Ground clutter and means of the Filter types are significantly different. But it is noted that there is no interaction between the Ground clutter and Filter types. Furthermore, since F-statistic indicates rejection of the null hypothesis, we therefore apply the LSD on Ground clutter to find out which means differ significantly. 39 | P a g e Multiple Comparisons LSD (I) Ground_clutter low (J) Ground_clutter medium high medium low 95% Confidence Interval Lower Upper bound bound -8.80 -1.70 -12.67 -5.58 Mean Difference (I-J) -5.25(*) -9.13(*) Std. Error 1.665 1.665 Sig. .007 .000 5.25(*) 1.665 .007 1.70 8.80 1.665 1.665 1.665 .034 .000 .034 -7.42 5.58 .33 -.33 12.67 7.42 high -3.88(*) high low 9.13(*) medium 3.88(*) * The mean difference is significant at the .05 level. Topic # 42: Least Significant Difference (L.S.D) Perform the analysis of variance on the following data and analyze the treatment means using the LSD test with a 0.05 level of significance. Treatments 1 2 Blocks3 4 1 2 3 4 5 6 1 1 3 2 3 4 6 3 6 4 7 2 4 8 8 3 3 5 4 2 2 1 3 1 (Int to Stat-theory, P-II) Ex # 20.7, Pg # 325 From the menus choose: Analyze General Linear Model Univariate … 40 | P a g e Multiple Comparisons LSD (I) Treatment 1 2 3 4 5 95% Confidence Interval Lowe Upper bound bound -4.14 -.36 -4.89 -1.11 Mean Difference (I-J) -2.25(*) -3.00(*) Std. Error .885 .885 Sig. .023 .004 4 -4.00(*) .885 .000 -5.89 5 -1.75 .885 .067 -3.64 .14 6 .00 .885 1.000 -1.89 1.89 1 2.25(*) .885 .023 .36 4.14 3 4 5 -.75 -1.75 .50 .885 .885 .885 .410 .067 .580 -2.64 -3.64 -1.39 1.14 .14 2.39 6 2.25(*) .885 .023 .36 4.14 1 (J) Treatment 2 3 -2.11 3.00(*) .885 .004 1.11 4.89 2 .75 .885 .410 -1.14 2.64 4 -1.00 .885 .276 -2.89 .89 5 6 1 1.25 3.00(*) 4.00(*) .885 .885 .885 .178 .004 .000 -.64 1.11 2.11 3.14 4.89 5.89 2 1.75 .885 .067 -.14 3.64 3 1.00 .885 .276 -.89 2.89 5 2.25(*) .885 .023 .36 4.14 6 4.00(*) .885 .000 2.11 5.89 1 2 3 1.75 -.50 -1.25 .885 .885 .885 .067 .580 .178 -.14 -2.39 -3.14 3.64 1.39 .64 4 -2.25(*) .885 .023 -4.14 -.36 6 1.75 .885 .067 -.14 3.64 1 .00 .885 1.000 -1.89 1.89 2 -2.25(*) .885 .023 -4.14 -.36 -3.00(*) -4.00(*) -1.75 * The mean difference is significant at the .05 level. .885 .885 .885 .004 .000 .067 -4.89 -5.89 -3.64 -1.11 -2.11 .14 6 3 4 5 41 | P a g e Topic # 43: Duncan’s Multiple Range Test Use Duncan’s multiple range test for the given data to compare all pairs of treatment means. Assume that α = 0.05 Treatments 1 2 Blocks3 4 1 2 3 4 5 6 1 1 3 2 3 4 6 3 6 4 7 2 4 8 8 3 3 5 4 2 2 1 3 1 (Int to Stat-theory, P-II) Ex # 20.8, Pg # 328 From the menus choose: Analyze General Linear Model Univariate … Values Duncan N Treatment 1 Subset 1 2 3 1 4 1.75 6 4 1.75 5 4 3.50 2 4 4.00 4.00 3 4 4.75 4.75 4 4 Sig. 3.50 5.75 .079 .200 .079 Means for groups in homogeneous subsets are displayed. The error term is Mean Square (Error) = 1.567. Alpha = .05. Topic # 44: Tukey’s Test To illustrate Tukey’s test, we use the data from the cotton weight percentage experiment in the given data. With α = 0.05 and f = 20 degree of freedom for error. Observed Tensile Length Weight of Cotton 1 2 3 4 5 15 7 7 15 11 9 20 12 17 12 18 18 25 14 18 18 19 19 30 19 25 22 19 23 35 7 10 11 15 11 (Design & experiment) Ex # 3.7, Pg # 97 42 | P a g e From the menus choose: Analyze General Linear Model Univariate … Multiple Comparisons (I) weights 15 20 25 30 35 Mean Difference (I-J) -5.60(*) -7.80(*) Std. Error 1.522 1.522 Sig. .015 .001 30 -11.80(*) 1.522 .000 -16.46 -7.14 35 -1.00 1.522 .963 -5.66 3.66 15 5.60(*) 1.522 .015 .94 10.26 25 30 35 -2.20 -6.20(*) 4.60 1.522 1.522 1.522 .609 .007 .054 -6.86 -10.86 -.06 2.46 -1.54 9.26 15 7.80(*) 1.522 .001 3.14 12.46 20 2.20 1.522 .609 -2.46 6.86 30 35 -4.00 6.80(*) 1.522 1.522 .112 .003 -8.66 2.14 .66 11.46 15 20 11.80(*) 6.20(*) 1.522 1.522 .000 .007 7.14 1.54 16.46 10.86 25 4.00 1.522 .112 -.66 8.66 35 10.80(*) 1.522 .000 6.14 15.46 15 1.00 1.522 .963 -3.66 5.66 20 -4.60 -6.80(*) -10.80(*) 1.522 1.522 1.522 .054 .003 .000 -9.26 -11.46 -15.46 .06 -2.14 -6.14 (J) weights 20 25 25 30 95% Confidence Interval Upper bound Lower bound -10.26 -.94 -12.46 -3.14 Topic # 45: Regression analysis of 23 Factorial Design A chemical engineer is investigating the yields of processes. Three process variables are of interest, temperature, and pressure and catalyst concentration. Each variable can be run at low and high level, and the engineer decides to run a 23 design with four centers points. 43 | P a g e Run 1 2 3 4 5 6 7 8 9 10 11 12 Temperature 120 160 120 160 120 160 120 160 140 140 140 140 Pressure 40 40 80 80 40 40 80 80 60 60 60 60 concentration 15.0 15.0 15.0 15.0 30.0 30.0 30.0 30.0 22.5 22.5 22.5 22.5 Yields 32 46 57 65 36 48 57 68 50 44 53 56 (Design & experiment) Ex # 10.2, Pg # 402 Note: First of all we will make the coded variables of the given variables temperature as 𝑋1 , pressure as 𝑋2 , concentration as 𝑋3 . By using the formula 𝑋1 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒−60 20 and 𝑋3 = 𝑐𝑜𝑛𝑐−22.5 From the menus choose: Transform Compute Variable … 7.5 . 𝑡𝑒𝑚𝑝−140 20 , 𝑋2 = Analyze Reports Case Summaries … 2^4 FACTORIAL DATA x1 x2 x3 yield 1 -1 -1 -1 32 2 1 -1 -1 46 3 -1 1 -1 57 4 1 1 -1 65 5 -1 -1 1 36 6 1 -1 1 48 7 -1 1 1 57 8 1 1 1 68 9 0 0 0 50 10 0 0 0 44 11 0 0 0 53 12 0 0 0 56 -1 -1 -1 32 Total 44 | P a g e From the menus choose: Analyze Regression Linear … Coefficients Unstandardized Coefficients Model 1 (Constant) x1 x2 x3 B 51.000 5.625 10.625 1.125 Std. Error .966 1.183 1.183 1.183 Standardized Coefficients Beta .449 .848 .090 t B 52.783 4.753 8.979 .951 Sig. Std. Error .000 .001 .000 .370 Topic # 46: Dennett’s Test (3T) To illustrate Dennett’s test, consider the experiment development engineer is interested in determining if the cotton weight % in a synthetic fiber affect the tensile strength and she has run a completely randomized experiment with 5 levels of cotton weight % and 5 replicates. Weight % of Cotton 15 20 25 30 35 1 7 12 14 19 7 Observed Tensile Strength 2 3 4 7 15 11 17 12 18 18 18 19 25 22 19 10 11 15 5 9 18 19 23 11 Perform the analysis of Dennett’s test at 5% level of significance. (Design & experiment) Ex # 3.10, Pg # 103 From the menus choose: Analyze General Linear Model Univariate … (post hoc test) 45 | P a g e Multiple Comparisons Dunnett’s T3 (I) weights 1 2 3 4 5 Mean Difference (I-J) -5.60 -7.80(*) Std. Error 2.049 1.761 Sig. .180 .026 Lower bound -13.11 -14.60 Upper bound 1.91 -1.00 4 -11.80(*) 1.897 .003 -18.86 -4.74 5 -1.00 1.970 1.000 -8.26 6.26 1 5.60 2.049 .180 -1.91 13.11 3 4 5 -2.20 -6.20 4.60 1.679 1.822 1.897 .845 .074 .273 -8.61 -12.93 -2.36 4.21 .53 11.56 1 7.80(*) 1.761 .026 1.00 14.60 2 2.20 1.679 .845 -4.21 8.61 4 -4.00 1.490 .197 -9.53 1.53 5 6.80(*) 11.80(*) 6.20 1.581 1.897 1.822 .025 .003 .074 .86 4.74 -.53 12.74 18.86 12.93 3 4.00 1.490 .197 -1.53 9.53 5 10.80(*) 1.732 .002 4.45 17.15 1 1.00 1.970 1.000 -6.26 8.26 2 -4.60 -6.80(*) -10.80(*) 1.897 1.581 1.732 .273 .025 .002 -11.56 -12.74 -17.15 2.36 -.86 -4.45 (J) weights 2 3 1 2 3 4 95% Confidence Interval Topic # 47: Test of Equality of Variance (Bartlett’s) We can apply the Bartlett’s test to the tensile strength data from the cotton weight percentage experiment in the following data. Weight % of Cotton 15 20 25 30 35 1 7 12 14 19 7 Observed Tensile Strength 2 3 4 7 15 11 17 12 18 18 18 19 25 22 19 10 11 15 5 9 18 19 23 11 (Design & experiment) Ex # 3.4, Pg # 82 From the menus choose: Analyze Compare Means One Way ANOVA … (option) 46 | P a g e Test of Homogeneity of Variances Levine Statistic .939 df1 df2 4 Sig. .462 20 Topic # 48: Effect Size for the One-Way (Eta-Square) Compute the Eta-Square for the following data shows version of the musical experiment of the five groups. 1 16 16 14 13 12 Groups 3 26 24 22 20 20 2 23 21 20 20 17 4 11 9 7 7 7 5 10 9 9 6 6 (Statistics bhvrl Scnc) Pg # 304 From the menus choose: Analyze General Linear Model Univariate … (option/effect size) Tests of Between-Subjects Effects Source Corrected Model Intercept Groups Type III Sum of Squares Mean Square df 884.400(a) 4 221.100 5329.000 1 5329.000 884.400 4 221.100 Error 85.600 20 4.280 Total 6299.000 25 970.000 24 Corrected Total F 51.659 1245.09 3 51.659 Sig. Partial Eta Squared .000 .912 .000 .984 .000 .912 Topic # 49: Control Chart by the Cases is Units A food company puts mango juice into cans advertised as containing 10 ounces, of the juice. The weights of the juice drained from cans immediately after filling for 20 samples are taken by random method. Construct control chart. 47 | P a g e Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Weight of each can ( 4 cans in each sample, n = 4) 15 12 13 20 10 8 8 14 8 15 17 10 12 17 11 12 18 13 15 4 20 16 14 20 15 19 23 17 13 23 14 16 9 8 18 5 6 10 24 20 5 12 20 15 3 15 18 18 6 18 12 10 12 9 15 18 15 15 6 16 18 17 8 15 13 16 5 4 10 20 8 10 5 15 10 12 6 14 33 14 (Txt Busi Stat) Ex # 4, Pg # 32 From the menus choose: Analyze Quality Control Control Charts … 48 | P a g e Topic # 50: Control Chart by the Sub-Groups The tables below gives the measurements obtain in 20 samples (subgroups). Constructs the control charts based on the mean and the range. The values of these statistics are given below for the respective samples. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -1 2 1 2 1 1 -1 1 2 -2 0 2 0 0 -1 1 2 2 0 3 2 0 1 1 -1 -1 1 1 1 1 1 1 1 0 2 -1 1 0 2 -3 1 1 0 0 0 2 0 2 -1 -2 -3 -1 -3 -1 1 2 -1 1 1 -1 0 0 0 -1 0 0 -2 -1 0 2 2 0 2 0 1 0 0 0 -1 1 1 1 1 0 1 1 0 1 1 2 -2 0 1 1 2 -1 -2 1 0 0 (Txt Busi Stat) Ex # 1, Pg # 28 49 | P a g e From the menus choose: Analyze Quality Control Control Charts … Topic # 51: Second Degree Parabola Fit a Second degree parabola to the following data, taking X as an independent variable. X Y 0 1 1 1.8 2 3 4 1.3 2.5 6.3 (Int to Stat-theory, P-I) Ex # 12.2, Pg # 489 50 | P a g e From the menus choose: Analyze Regression Curve Estimation … Model Summary and Parameter Estimates Model Summary Equation Quadratic R Square .915 F Parameter Estimates df1 10.735 df2 2 Sig. 2 1 1.6 2 4.5 3 13.8 4 40.2 B1 1.420 .085 Topic # 52: Exponential Curve Fit an Exponential curve 𝑌 = 𝑎𝑒 𝑏𝑋 to the following data. X Y Constant b2 -1.070 .550 5 6 125.0 363.0 (Int to Stat-theory, P-I) Ex # 12.5, Pg # 495 From the menus choose: Analyze Regression Curve Estimation … Model Summary and Parameter Estimates Model Summary Equation Exponential R Square 1.000 F 32439.854 df1 Parameter Estimates df2 1 Sig. .000 4 Constant .524 b1 1.090 Topic # 53: Third Degree Parabola Fit a Third degree parabola to the following data, taking X as an independent variable. X Y 0 1 1 1.8 2 1.3 3 2.5 4 6.3 (Int to Stat-theory, P-I) Prb # 12.12, Pg # 506 From the menus choose: Analyze Regression Curve Estimation … 51 | P a g e Model Summary and Parameter Estimates Model Summary Equatio n Cubic R Square F .997 df1 98.016 Parameter Estimates df2 3 Constan t Sig. 1 1.030 .074 b1 b2 1.725 b3 -1.400 .325 Topic # 54: Power Curve Fit the curve of the from Y = 𝑎𝑋 𝑏 to the following data on the unit cost in dollars of producing certain electronic component’s and the number of units produced Lot size (X) 50 100 250 500 1000 Unit cost (Y) 108 53 24 9 5 Use the result to estimate the unit cost for a lot of 400 components. (Int to Stat-theory, P-I) Prb # 12.21, Pg # 507 From the menus choose: Analyze Regression Curve Estimation … Model Summary and Parameter Estimates Equation Model Summary R Square Power .994 F 530.639 df1 Parameter Estimates df2 1 Sig. 3 .000 Constant 6480.770 b1 -1.040 Case Summaries 1 2 3 4 5 6 Total N lot size 50 100 250 500 1000 400 6 cost 108 53 24 9 5 5 Fit for cost with lot size from CURVEFIT, MOD_2 POWER 110.86627 53.91955 20.79277 10.11252 4.91820 12.75381 6 52 | P a g e Topic # 55: Testing Hypothesis about P’s of The Multinomial Distribution Two hundred were chosen at random from a set of table. The frequencies of the digits were. Digits Frequency 0 18 1 19 2 23 3 21 4 16 5 25 6 22 7 20 8 21 9 15 (Int to Stat-theory, P-II) Ex # 17.8, Pg # 192 From the menus choose: Data Weight cases … Analyze Non-Parametric Tests Chi-Square … Test Statistics Digits ChiSquare(a) df Asymp. Sig. 4.300 9 .891 Topic # 56: Goodness of Fit Test Five pennies were tossed 1000 times and the number of heads observed as given below Number of heads Frequency 0 38 1 144 2 342 3 287 4 164 5 25 Test whether a Binomial distribution given as satisfactory fit to this data while the p = 0.494. (Int to Stat-theory, P-II) Ex # 17.10, Pg # 196 From the menus choose: Transform Compute Variable  Make a target variable “Expected” and fit Binomial distribution. Data Weight cases … Analyze Non-Parametric Tests Chi-Square …  Values (Add all values from the expected column) 53 | P a g e Num_of_Heads 0 Observed N 38 Expected N 33.2 1 144 161.9 -17.9 2 342 316.2 25.8 3 287 308.7 -21.7 4 164 150.7 13.3 25 29.4 -4.4 5 Total Residual 4.8 1000 Test Statistics Num_of_Heads ChiSquare(a) df 8.146 5 Asymp. Sig. .148 Topic # 57: Testing Hypothesis about Equality of Several Proportion Form the adult male population of seven large cities random samples of sizes indicated below were taken, and the numbers of married and single man recorded. City Married Single Total A 133 36 169 B 164 57 221 C 155 40 195 D 106 37 143 E 153 55 208 F 123 33 156 G 146 36 182 Total 980 294 1274 (Int to Stat-theory, P-II) Ex # 17.21, Pg # 218 From the menus choose: Data Weight cases … Analyze Descriptive Statistics Crosstabs … 54 | P a g e Chi-Square Tests Pearson Chi-Square Likelihood Ratio Linear-by-Linear Association N of Valid Cases Value 5.337 5.327 6 6 Asymp. Sig. (2-sided) .501 .503 1 .617 df .250 1274 Topic # 58: The Chi-Square Test as Test of Homogeneity In certain community, a random sample of 50 men and another sample of 50 women over 21 year of age were asked about their eruptional background, classified as junior high, senior high or college. The results are, Junior High 13 23 Male Female Senior High 25 20 College 12 7 (Int to Stat-theory, P-II) Ex # 17.22, Pg # 220 From the menus choose: Data Weight cases … Analyze Descriptive Statistics Crosstabs … Chi-Square Tests Pearson Chi-Square Likelihood Ratio Linear-by-Linear Association N of Valid Cases Value 4.649 4.703 Asymp. Sig. (2-sided) df 4.275 2 2 .098 .095 1 .039 100 Topic # 59: Fit the Second Degree Curve Trend Fit the second degree trend curve (parabola) to the following data and compute the trend values. Year Index (whole sale price) Coded 1931 96 -7 1933 87 -5 1935 91 -3 1937 102 -1 1939 108 1 1941 139 3 1943 307 5 1945 289 7 (Int to Stat-theory, P-I) Ex # 13.7, Pg # 527 55 | P a g e From the menus choose: Analyze Regression Curve Estimation … Model Summary and Parameter Estimates Model Summary Equation Quadratic R Square .874 F df1 17.314 Parameter Estimates df2 2 Sig. 5 .006 Constant 110.219 b1 15.482 b2 2.007 Topic # 60: Growth Model In the exercise 3.22 we presented data on U.S. real GDP for the period 1972-1991 suppose we want to find out the rate of growth of real GDP in this period. Let 𝑌𝑡 = real GDP (RGDP) at time t and 𝑌0 = the initial (1972) value of real GDP. Time 1972 1973 1974 1975 1976 1977 1978 GDP(1987) 3,107.1 3,268.6 3,248.1 3,221.7 3,380.8 3,533.3 3,703.5 Time 1979 1980 1981 1982 1983 1984 1985 GDP(1987) 3,796.8 3,776.3 3,843.1 3,760.3 3,906.6 4,148.5 4,279.8 Time 1986 1987 1988 1989 1990 1991 GDP(1987) 4,404.5 4,539.9 4,718.6 4,838.0 4,877.5 4,821.0 (Basic Eco) Sec # 6.5, Pg # 169 From the menus choose: Analyze Regression Curve Estimation … Model Summary and Parameter Estimates Model Summary Equation Growth R Square .974 F 668.959 df1 Parameter Estimates df2 1 18 Sig. .000 Constant 8.014 b1 .025 Topic # 61: Autocorrelation Function for 25 Lags (ACF) Calculate the autocorrelation function for the GDP of United States, 1970 – I to 1991 - IV by using the 25 lags from the following data. 56 | P a g e Quarter 1970 – I 1970 – II 1970 – III 1970 – IV 1971 – I 1971 – II 1971 – III 1971 – IV 1972 – I 1972 – II 1972 – III 1972 – IV 1973 – I 1973 – II 1973 – III 1973 – IV 1974 – I 1974 – II 1974 – III 1974 – IV 1975 – I 1975 – II GDP 2872.8 2860.3 2896.6 2873.7 2942.9 2947.4 2966.0 2980.8 3037.3 3089.7 3125.8 3175.5 3253.3 3267.6 3264.3 3289.1 3259.4 3267.6 3239.1 3226.4 3154.0 3190.4 Quarter 1975 – III 1975 – IV 1976 – I 1976 – I1 1976 – I1I 1976 – IV 1977 – I 1977 – II 1977 – III 1977 – IV 1978 – I 1978 – II 1978 – III 1978 – IV 1979 – I 1979 – II 1979 – III 1979 – IV 1980 – I 1980 – II 1980 – III 1980 – IV GDP 3249.9 3292.5 3356.7 3369.2 3381.0 3416.3 3466.4 3525.0 3574.4 3567.2 3591.8 3707.0 3735.6 3779.6 3780.8 3784.3 3807.5 3814.6 3830.8 3732.6 3733.5 3808.5 Quarter 1981 – I 1981 – II 1981 – III 1981 – IV 1982 – I 1982 – II 1982 – III 1982 – IV 1983 – I 1983 – II 1983 – III 1983 – IV 1984 – I 1984 – II 1984 – III 1984 – IV 1985 – I 1985 – II 1985 – III 1985 – IV 1986 – I 1986 – II GDP 3860.5 3844.4 3864.5 3803.1 3756.1 3771.1 3754.4 3759.6 3783.5 3886.5 3944.4 4012.1 4089.5 4144.0 4166.4 4194.2 4221.8 4254.8 4309.0 4333.5 4390.5 4387.7 Quarter 1986 – III 1986 – IV 1987 – I 1987 – II 1987 – III 1987 – IV 1988 – I 1988 – II 1988 – III 1988 – IV 1989 – I 1989 – II 1989 – III 1989 – IV 1990 – I 1990 – II 1990 – III 1990 – IV 1991 – I 1991 – II 1991 – III 1991 – IV GDP 4412.6 4427.1 4460.0 4515.3 4559.3 4625.5 4655.3 4704.8 4734.5 4779.7 4809.8 4832.4 4845.6 4859.7 4880.8 4900.3 4903.3 4855.1 4824.0 4840.7 4862.7 4868.0 (Basic Eco) Sec # 21.3, Pg # 716 From the menus choose: Analyze Time Series Autocorrelations … 57 | P a g e Autocorrelations Autocorrel ation Lag 1 Value .969 Std. Error(a) df Box-Ljung Statistic Sig.(b) Value df .105 85.462 1 .000 2 .935 .104 166.016 2 .000 3 .901 .104 241.717 3 .000 4 .866 .103 312.393 4 .000 5 .830 .102 378.100 5 .000 6 .791 .102 438.566 6 .000 7 .752 .101 493.849 7 .000 8 .713 .101 544.108 8 .000 9 .675 .100 589.773 9 .000 10 .638 .099 631.121 10 .000 11 .601 .099 668.328 11 .000 12 .565 .098 701.649 12 .000 13 .532 .097 731.556 13 .000 14 .500 .097 758.290 14 .000 15 .468 .096 782.020 15 .000 16 .437 .095 803.025 16 .000 17 .405 .095 821.346 17 .000 18 .375 .094 837.237 18 .000 19 .344 .093 850.786 19 .000 20 .313 .093 862.172 20 .000 21 .279 .092 871.390 21 .000 22 .246 .091 878.647 22 .000 23 .214 .091 884.216 23 .000 24 .182 .090 888.314 24 .000 25 .153 .089 891.246 25 .000 Topic # 62: Partial Autocorrelation Function for 25 Lags (PACF) Calculate the partial autocorrelation function for the GDP of United States, 1970 – I to 1991 - IV by using the 25 lags from the following data. 58 | P a g e Quarter 1970 – I 1970 – II 1970 – III 1970 – IV 1971 – I 1971 – II 1971 – III 1971 – IV 1972 – I 1972 – II 1972 – III 1972 – IV 1973 – I 1973 – II 1973 – III 1973 – IV 1974 – I 1974 – II 1974 – III 1974 – IV 1975 – I 1975 – II GDP 2872.8 2860.3 2896.6 2873.7 2942.9 2947.4 2966.0 2980.8 3037.3 3089.7 3125.8 3175.5 3253.3 3267.6 3264.3 3289.1 3259.4 3267.6 3239.1 3226.4 3154.0 3190.4 Quarter 1975 – III 1975 – IV 1976 – I 1976 – I1 1976 – I1I 1976 – IV 1977 – I 1977 – II 1977 – III 1977 – IV 1978 – I 1978 – II 1978 – III 1978 – IV 1979 – I 1979 – II 1979 – III 1979 – IV 1980 – I 1980 – II 1980 – III 1980 – IV GDP 3249.9 3292.5 3356.7 3369.2 3381.0 3416.3 3466.4 3525.0 3574.4 3567.2 3591.8 3707.0 3735.6 3779.6 3780.8 3784.3 3807.5 3814.6 3830.8 3732.6 3733.5 3808.5 Quarter 1981 – I 1981 – II 1981 – III 1981 – IV 1982 – I 1982 – II 1982 – III 1982 – IV 1983 – I 1983 – II 1983 – III 1983 – IV 1984 – I 1984 – II 1984 – III 1984 – IV 1985 – I 1985 – II 1985 – III 1985 – IV 1986 – I 1986 – II GDP 3860.5 3844.4 3864.5 3803.1 3756.1 3771.1 3754.4 3759.6 3783.5 3886.5 3944.4 4012.1 4089.5 4144.0 4166.4 4194.2 4221.8 4254.8 4309.0 4333.5 4390.5 4387.7 Quarter 1986 – III 1986 – IV 1987 – I 1987 – II 1987 – III 1987 – IV 1988 – I 1988 – II 1988 – III 1988 – IV 1989 – I 1989 – II 1989 – III 1989 – IV 1990 – I 1990 – II 1990 – III 1990 – IV 1991 – I 1991 – II 1991 – III 1991 – IV GDP 4412.6 4427.1 4460.0 4515.3 4559.3 4625.5 4655.3 4704.8 4734.5 4779.7 4809.8 4832.4 4845.6 4859.7 4880.8 4900.3 4903.3 4855.1 4824.0 4840.7 4862.7 4868.0 (Basic Eco) Sec # 21.3, Pg # 716 From the menus choose: Analyze Time Series Autocorrelations … 59 | P a g e Partial Autocorrelations Partial Autocorrel ation .969 Std. Error .107 2 -.058 .107 3 -.020 .107 4 -.045 .107 5 -.024 .107 6 -.062 .107 7 -.029 .107 8 -.024 .107 9 .009 .107 10 -.010 .107 11 -.020 .107 12 -.012 .107 13 .020 .107 14 -.012 .107 15 -.021 .107 16 -.001 .107 17 -.041 .107 18 -.005 .107 19 -.038 .107 20 -.017 .107 21 -.066 .107 22 -.019 .107 23 -.008 .107 24 -.018 .107 25 .017 .107 Lag 1 Topic # 63: Time Series Graph Plot the time series graph for the four quarters of each year of 1983 to 1986. Data is given the following table. Year 1983 1984 1985 1986 I 20 15 15 18 Quarter II III 26 43 21 43 32 45 36 50 IV 57 68 73 78 (Stat for Eco & Mngmnt) Ex # 14.1, Pg # 352 From the menus choose: Analyze Time Series Sequence Charts … 60 | P a g e 80 values 60 40 20 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 quarter Topic # 64: Exponential Smoothing The data from 1985 January to 1987 December the total 36 observations are displayed in the following table; Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1985 1.3 .8 1.2 .9 .9 1.1 .8 .9 .9 1.1 .9 1.3 1986 1.0 1.0 1.1 1.6 1.1 1.0 1.2 1.0 .9 1.0 1.0 1.1 1987 .8 1.0 1.0 1.2 1.1 1.3 1.1 1.1 1.3 1.2 1.3 1.1 (Busi Stat) Sec # 7.6.1, pg # 154 From the menus choose: Transform Create Time Series … (smoothing) Analyze Time Series Sequence Charts … 61 | P a g e THE END 62 | P a g e