MGT223 1 Material Provided: Statistical Tables, Appendix of SPSS output Material to be introduced: Candidates may take into the examination an A4 sheet of personal notes which must be attached to the script at the end of the examination. SOLUTIONS INCLUDED MGT223 1. Historically it has taken a local council an average of 14 weeks to reach a decision about planning applications for building development work submitted by construction companies. In an attempt to reduce this time a new system for processing such applications has been introduced. The first 25 applications to be processed under this new system took an average of 12 weeks with a standard deviation of 4.3 weeks. Comment on the efficacy of the new system. (12) Null -H0: (no change) Alternative - H1: (has new system had the desired effect?) 1 mark n =25 ̅ = 12 s = 4.3 1 mark Assume normal population 1 mark 95% CI: 3 marks The CI for the average age ranges 10.2 to 13.8 weeks – the data are not consistent with the null hypothesis of the average being 14 – how strong is the evidence? 1 mark p-value = P(̅ ≤ 12|H0) t24,0.025 = 2.0639 t24,0.01 = 2.4922 3 marks Good evidence against the null hypothesis. Based on the sample evidence there is good evidence against average processing time having stayed the same. The average processing time appears to have fallen to somewhere in the region 10.2 to 13.8 weeks 2 marks (12) Deduct 4 if z instead of t 7750.112 86.00639.212 25 3.4 12 025.0,24 025.0,1 t n s tx n 025.001.0 )33.2( ) 25 3.4 1412 ( valuep tP tP 2. A manufacturer claims that only 1% of items produced are defective. A sample of 40 items was taken and three defective items were found. Assess the manufacturer’s claim. (12) Null - H0: Alternative - H1: (manufacturer’s claim is false – more than 1% defective) 1 mark n =40 3 defectives, p = 0.075 1 mark The sample size is large (n > 20) so use approximation to Binomial np = 3, n =4 (both < 5 so use charts for CI and Poisson approximation = n for p-value) 2 mark 95% CI for true proportion (from charts): 0.018 to 0.200. Wide but does not include the hypothesised value of 1%. Undertake a hypothesis test to assess the strength of the evidence 3 mark p-value = Pr(number defects ≥ 3| H0: = Pr(number of defects ≥ 3| Binomial n=40 = Pr(number of defects ≥ 3| Poisson, = n = 0.0079 3 mark There is good evidence against the null hypothesis. While we can’t say what the true value is with great precision there is evidence that it is greater than 1%. The best estimate of the defect rate is between 1.8 and 20%. 2 mark 3. Two processes are being considered by a manufacturer who has obtained some data relating to production per hour. The data were analysed using SPSS and the resultant output is provided in the Appendix. a) Is there any evidence to suggest that the two processes differ in terms of their average productivity? State, and assess where possible, any assumptions you have made in arriving at your conclusion. (9) Independent samples t-test: The test assumes normal distributions, data at least interval and equal variances. Check normality via the Explore output: Values for skewness and kurtosis provide no evidence against normality Process A: .252/.564=0.45, 1.492/1.091=1.36 – both less than 2 Process B: .497/.501=0.99, .310//1.091=0.32 – both less than 2 This is backed up by p-values from Shapiro-Wilk tests (0.133 and 0.276) The Normal probability plots look to be fairly linear. The boxplots seem fairly symmetrical. 3 marks The test also assumes equal variances which we can check with Levene’s test H0: H1: p = 0.016 - evidence against equal variances – we cannot assume equal variances so should follow the bottom row of table. 2 marks Main test H0: 2 H1: ≠ 2 (do the processes differ?) 1 mark 95% CI for difference (-4.772 to -.221). This does not include 0 so the data are not consistent with the null hypothesis. It looks as though process B has a higher average. Looking at the p-value 0.033 – some evidence against the null hypothesis. The results suggest that there may be a difference with process B giving a higher average output – the difference could be almost 5 units on average but could be much smaller 0.2 units per hour) 3 marks b) It is recognised that production figures may vary from hour to hour, however a wide range in hourly production figures is not desirable. What does the analysis tell you about the relative variability of production? (2) Levene’s test reported above indicates good evidence against equal variances (p=0.016). Looking at the standard deviation we can see that Process B has the lower standard deviation of 2.49 compared with 3.84 for Process A. 2 marks c) Which process would you recommend and why? (3) Process B – there is evidence of higher average output (up to 5 units per hour) and there is less variability in output. 3 marks 4. During times of business decline many groups offer suggestions for spurring the economy into a turnaround. A survey was conducted among 100 business executives, 100 economists and 100 government officials to find the opinion of each regarding the best way of reversing the trend of business decline. Their responses are tabulated below: Group Opinion Business executives Economists Government officials Increase government spending 10 15 39 Cut personal income taxes 37 37 33 Decrease interest rates 24 34 15 Offer tax incentives to business 29 14 13 Cross-tabulation of favoured opinion by group. The data were analysed in SPSS. The resultant output is given in the appendix. a) Interpret this output. Looking at the data it appears that all groups advocate the cutting of personal taxes to some degree but may differ with respect to the other options. (1) Null -H0: No association between group and preferred solution. Alternative - H1: Different groups advocate different solutions. (1) Chi-Square = 38.86, p < 0.0005 – extremely strong evidence against null. There does appear to be an association. (2) The large standardised residuals indicate that there are differences between the 3 groups. Business executives favour tax incentives, Economists favour decreasing interest rates and Government officials favour Increasing government spending. (2) (b) Write a paragraph summarising your findings suitable for inclusion in a management report. There are clear differences between each group (p < 0.0005). There are similar proportions (around 35% of each group) who give cutting personal taxes as their preferred option but the other three options are favoured differently by different groups. 29% of Business Execs state tax incentives (compared with only 14% of economists and 13% of Government Officials). Economists favour decreased interest rates (34% compared with 24% of Business Execs and 15% of Government Officials). Increased government spending has the support of 39% of Government Officials compared with 15% of Economists and only 10% of Business Execs. (4) 5. What statistical method or model would you use to analyse the following situations? In each case give brief details of how the modelling would be undertaken. a) An electronics company is reviewing its assembly-line procedures. It has been established that a particular task takes 15.5 minutes to complete. A random sample of 9 employees has been taught a new method. After the training period each employee was timed performing the task. A single sample t-test (with CI). Assume normaility. Null hypothesis: = 15.5 Alternative hypothesis: < 15.5 (an improvement) (2) b) An organisation is trying to evaluate which of two computer systems to standardise on within the organisation. In terms of price, specification and performance there is little to choose between the two models under consideration. However the IT manager is concerned about the possible maintenance and repair costs. Data are available comprising the mean and standard deviation of annual maintenance costs for samples of 20 of each model. Independent samples t-test (with CI for difference in means). Null hypothesis: = Alternative hypothesis: ≠ (a difference) (2) Might also be interested in variability and need to test this anyway for the independent samples t-test. Use F test Null hypothesis: = Alternative hypothesis: ≠ (a difference) (2) c) A marketing experiment has been conducted to compare five different types of packaging for a laundry detergent. Each package was shown to 40 different potential consumers who were asked to rate the attractiveness of the product on a scale of 1 to 10. One way ANOVA to test for equality of means across the five types of packaging. Null hypothesis: = = = = Alternative hypothesis: not all equal (2) d) The manager of a local health and leisure club is investigating the frequency of its use by members. She has collected data on a random sample of members comprising their weekly attendance, distance from home to the club, number of children in family and annual income. How might this data be used to forecast attendance for other members? Regression analysis with dependent variable of attendance and distance from home, number of children and annual income as explanatory variables. Assess effect of these variables via p-values and r-squared. (2) 6. A university department is interested in the variability of grades between modules. Of particular interest are two core modules, MOD102 and MOD104. Data from the most recent cohort of 28 students has been analysed using SPSS and the results are given in the appendix. a) Provide a full interpretation of these results, indicating the assumptions upon which each test is based and whether or not you believe them to be reasonable. (10) Paired t-test Assumes normal distribution (of differences), data at least interval and paired data. (1) Check normality via the Explore output: Values for skewness and kurtosis provide no evidence against normality 0.102/.441=0.45, 0.098/.858=0.11 – both less than 2 This is backed up by p-values from Shapiro-Wilk test (0.720) The histogram seems fairly symmetrical. The Normal probability plot looks to be fairly linear. (2) Tests whether the average difference between the module grades could be 0 or is there evidence of a difference (2-sided since no information on which should be higher). Incorporates information on pairing so is more powerful if there are systematic differences between the pairs (likely to be the case here as we are looking at grades for each student). d d≠ (1) Correlation indicates efficacy of pairing. There is some weak evidence of positive correlation between the two sets of grades (r = 0.364, p=0.057) – grades do vary by student. By recognising the structure and pairing the data we can eliminate this variability and concentrate on the differences between the two modules. (2) The 95% CI for d Ranges from (-0.46 to 5.49). This includes 0 but at the lower end. There may be no difference but Module 2 may result in a higher average grade of up to 5 marks. Looking at the p-value =0.094, there is weak evidence against the null hypothesis. There is weak evidence to suggest a difference between the grades on the two modules. Module 2 may result in higher grades – the difference may be negligible but may be more than 5 marks on average. (2) Wilcoxon Signed Ranks Test. H0: samples drawn from sample population distribution H1: samples drawn from different population distributions Data need to be at least interval –as we are calculating differences between them. Also need to be paired - by student. Calculates differences between pairs and ranks them (ignoring signs). Under null the average of the positive ranks should be similar to the average of the negative ranks. Here there are 18 positive ranks (mean = 15.17) and 10 negative (mean=13.30). p-value = 0.111 - no evidence against null hypothesis. The sample of 28 students provides no clear evidence against the null. (2) b) Write a brief report, suitable for a non-statistician summarising the conclusions that may be drawn from the analysis. There is weak evidence (p=0.094) to suggest that the average grades could be higher for Module 2. The difference could be negligible but could be over 5 marks on average. (2) 7. A study is conducted to compare the job-satisfaction levels of assembly-line employees whose working environments are structured to different degrees. Also of interest is the relationship between length of employment and job satisfaction. The researchers wish to study the interaction between length of employment and the extent to which the working environment is structured and the effect of this interaction on job satisfaction. The data are given below (the higher the score the greater the job satisfaction): Nature of working environment Length of employment (years) Highly structured Moderately structured Unstructured 12 10 8 15 10 7 < 5 15 9 7 14 10 8 12 9 6 12 10 10 14 10 11 5 - 10 12 12 12 10 12 10 11 10 14 9 10 12 10 11 14 11 or more 9 10 15 9 10 15 10 12 18 Job satisfaction by length of employment and nature of working environment. An extract from an SPSS analysis of the data is given in the appendix a) Provide a full interpretation of the output. (8) Assume normal populations and homogeneity of variance. Normality is difficult to assess but there is no evidence against equal variances (p=0.250). (2) Interaction H0: no interaction between length of service and working environment H1: some interaction between length of service and working environment p-value < 0.0005 – extremely strong evidence of an interaction (1) Main effects H0: no length of service effect H1: some length of service effect p-value = 0.009 – good evidence against null – LoS does appear to affect job satisfaction. (1) H0: no working environment effect H1: some working environment effect p-value = 0.037 – some evidence against null – working environment does seem to have an impact on job satisfaction. (1) So main conclusions are that there is evidence that satisfaction varies according to length of service and working environment and extremely strong evidence of an interaction effect between length of service and working environment. (1) Marginal means give estimates of satisfaction levels for individual factors and interactions. It appears that job satisfaction tends to increase with length of service and those in a moderately structured environment tend to have lower satisfaction levels overall but there is also an interaction effect between the two factors which is best explored through the profile plots. (1) The profile plot shows that those employed for less than 5 years tend to be less satisfied generally unless they are working in a highly structured environment. Those employed between 5-10 years seem much less affected by the working environment and those employed for over 10 years are much more satisfied in an unstructured environment. (1) b) Write a paragraph describing, in non-technical terms, the main findings of this study. The data were analysed using two way analysis of variance. There was extremely strong evidence of an interaction between length of service and working environment on job satisfaction. There was good evidence of an additional length of service effect and some evidence of an additional working environment effect. (2) Those employed for less than 5 years tend to be less satisfied generally unless they are working in a highly structured environment. Those employed between 5-10 years seem much less affected by the working environment while those employed for over 10 years are much more satisfied in an unstructured environment. (2) 8. A study investigated the relationship between audit delays (Delay), the length of time (in days) from a company’s fiscal year end to the date of the auditor’s report, and variables that describe the client and the auditor: Industry: A dummy variable coded 1 for an industrial company and 0 otherwise. Public: A dummy variable coded 1 if the company was traded on the stock exchange. Quality: A measure of the overall quality of internal controls, as judged by the auditor, ranging from ‘virtually none’ (1) to ‘excellent‘ (5). Finished: A measure ranging from 1 to 4 as judged by the auditor, where 1 indicates ‘all work performed subsequent to the year-end’ and 4 indicates ‘most work performed prior to the year-end’. A sample of 40 companies provided data which was analysed using the SPSS package. The results of this analysis may be found in the appendix. a) Provide a full interpretation of the output. (14) The matrix scatterplots provide an overview of the data and allow us to identify potential relationships. DELAY appears to be linked to all four variables: INDUSTRY – More delay for industrial company (INDUSTRY = 1) PUBLIC – More delay for non public companies (PUBLIC = 0) QUALITY – Higher quality less delay FINISHED – Higher values of FINISHED – less delay (2) The correlations reinforce this: High negative correlation between FINISHED and DELAY (r = -0.624, p = 0.000). Less statistically significant correlations between DELAY and other 3 variables. (DELAY/INDUSTRY r = 0.287, p = 0.073) (DELAY/QUALITY r = -0.264, p = 0.100) (DELAY/PUBLIC r = -0.180, p = 0.265) (2) Regression1 DELAY = 88.78 + 8.25INDUSTRY – 1.76PUBLIC – 2.31QUALITY – 7.68FINISHED (r-square = 0.562, adj r-square = 0.512) - moderate explanatory power – 56% of variability in DELAY is accounted for.. (2) Coefficients of PUBLIC is not significant (p=0.590) – consider dropping from the model to give Regression 2: (1) Regression2 DELAY = 88.53 + 8.51INDUSTRY – 2.38QUALITY – 7.72FINISHED (r-square = 0.559, adj r-square = 0.522) The explanatory power is at the same level as for the previous regression so no explanatory power has been lost through the omission of PUBLIC. And adj r-square has increased. (1) All coefficients are statistically significant suggesting that we cannot remove these variables without affecting the model adversely. (1) Coefficient of INDUSTRY suggests that companies operating in this sector experience greater delays (by 8.5 days on average) Coefficient of QUALITY suggests that companies with higher quality of internal control experience fewer delays (2.4 days less on average). Coefficient of FINISHED suggests the companies where more work is performed prior to the year-end experience fewer delays (7.7 days less on average). (3) The most influential of the three variables is FINISHED with a Beta coefficient of -0.612. (1) The predictive power is moderate – around 56% of the variability in delays has been explained, leaving 44% unexplained. (1) b) Write a brief report, suitable for a non-statistician, summarising your findings and outlining the limitations of the analysis. We have used regression analysis to investigate the relationship between audit delays and sector, stock exchange listing, internal controls and preparedness. (1) The following equation was derived: DELAY = 88.53 + 8.51INDUSTRY – 2.38QUALITY – 7.72FINISHED (1) There is strong evidence that companies operating in Industry experience greater delays (8.5 days on average). There is good evidence that companies with higher quality internal controls experience fewer delays (2.4 days on average) as do those who are more prepared (extremely strong evidence of the influence of this factor – 7.7 day reduction on average). (1) The predictive power of this equation is fair - around 56% of the variability in delays can be explained by the equation. (1) End of Question Paper MGT223 MGT223 BUSINESS STATISTICS MGT223 Appendix of SPSS Output for questions 3, 4, 6, 7 & 8 MGT223 MGT223 QUESTION 3 - SPSS OUTPUT (page 1 of 2) Explore Descriptives PROCESS Statistic Std. Error OUTPUT2 Process A Mean 107.8706 .95965 95% Confidence Interval for Mean Lower Bound 105.8252 Upper Bound 109.9161 5% Trimmed Mean 107.9068 Median 108.5300 Variance 14.735 Std. Deviation 3.83858 Minimum 102.02 Maximum 113.07 Range 11.05 Interquartile Range 7.53 Skewness -.252 .564 Kurtosis -1.492 1.091 Process B Mean 110.3671 .54439 95% Confidence Interval for Mean Lower Bound 109.2316 Upper Bound 111.5027 5% Trimmed Mean 110.3111 Median 109.9900 Variance 6.224 Std. Deviation 2.49472 Minimum 106.60 Maximum 115.16 Range 8.56 Interquartile Range 3.03 Skewness .497 .501 Kurtosis -.310 .972 Tests of Normality PROCESS Kolmogorov-Smirnova Shapiro-Wilk Statistic df Sig. Statistic df Sig. OUTPUT2 Process A .139 16 .200* .914 16 .133 Process B .116 21 .200* .945 21 .276 *. This is a lower bound of the true significance. a. Lilliefors Significance Correction MGT223 QUESTION 3 - SPSS OUTPUT (page 2 of 2) Histograms Normal Q-Q Plots T-Test Group Statistics PROCESS N Mean Std. Deviation Std. Error Mean OUTPUT2 Process A 16 107.8706 3.83858 .95965 Process B 21 110.3671 2.49472 .54439 Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper OUTPUT2 Equal variances assumed 6.464 .016 -2.395 35 .022 -2.49652 1.04260 -4.61311 -.37993 Equal variances not assumed -2.263 24.319 .033 -2.49652 1.10331 -4.77205 -.22099 MGT223 QUESTION 4 – SPSS OUTPUT (page 1 of 1) Group * Opinion Crosstabulation 10 37 24 29 100 21.3 35.7 24.3 18.7 100.0 10.0% 37.0% 24.0% 29.0% 100.0% -2.5 .2 -.1 2.4 15 37 34 14 100 21.3 35.7 24.3 18.7 100.0 15.0% 37.0% 34.0% 14.0% 100.0% -1.4 .2 2.0 -1.1 39 33 15 13 100 21.3 35.7 24.3 18.7 100.0 39.0% 33.0% 15.0% 13.0% 100.0% 3.8 -.4 -1.9 -1.3 64 107 73 56 300 64.0 107.0 73.0 56.0 300.0 21.3% 35.7% 24.3% 18.7% 100.0% Count Expected Count % within Group Std. Residual Count Expected Count % within Group Std. Residual Count Expected Count % within Group Std. Residual Count Expected Count % within Group Business Executives Economists Government officials Group Total Increase government spending Cut personal income taxes Decrease interest rates Offer tax incentives to business Opinion Total Chi-Square Tests 38.862a 6 .000 37.301 6 .000 23.464 1 .000 300 Pearson Chi-Square Likelihood Ratio Linear-by-Linear Association N of Valid Cases Value df Asymp. Sig. (2-sided) 0 cells (.0%) have expected count less than 5. The minimum expected count is 18.67. a. MGT223 MGT223 QUESTION 6 – SPSS OUTPUT (page 1 of 2) Explore Descriptives Statistic Std. Error DIFFERENCE Mean 2.5189 1.44974 95% Confidence Interval for Mean Lower Bound -.4557 Upper Bound 5.4935 5% Trimmed Mean 2.4397 Median 3.2050 Variance 58.849 Std. Deviation 7.67128 Minimum -13.33 Maximum 21.44 Range 34.77 Interquartile Range 13.37 Skewness .102 .441 Kurtosis .098 .858 Tests of Normality Kolmogorov-Smirnova Shapiro-Wilk Statistic df Sig. Statistic df Sig. DIFFERENCE .100 28 .200* .975 28 .720 *. This is a lower bound of the true significance. a. Lilliefors Significance Correction DIFFERENCE MGT223 MGT223 QUESTION 6 – SPSS OUTPUT (page 2 of 2) T-Test Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 MOD102 83.3821 28 7.18082 1.35705 MOD104 80.8632 28 6.36915 1.20366 Paired Samples Correlations N Correlation Sig. Pair 1 MOD102 & MOD104 28 .364 .057 Paired Samples Test Paired Differences t df Sig. (2-tailed) Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the Difference Lower Upper MOD102 - MOD104 2.51893 7.67128 1.44974 -.45568 5.49354 1.738 27 .094 NPar Tests Wilcoxon Signed Ranks Test Ranks N Mean Rank Sum of Ranks MOD102 - MOD104 Negative Ranks 10a 13.30 133.00 Positive Ranks 18b 15.17 273.00 Ties 0c Total 28 a. MOD102 < MOD104 b. MOD102 > MOD104 c. MOD102 = MOD104 Test Statisticsa MOD102 - MOD104 Z -1.594b Asymp. Sig. (2-tailed) .111 a. Wilcoxon Signed Ranks Test b. Based on negative ranks. MGT223 MGT223 QUESTION 7 – SPSS OUTPUT (page 1 of 2) Levene's Test of Equality of Error Variancesa Dependent Variable: Job satisfaction F df1 df2 Sig. 1.352 8 36 .250 Tests the null hypothesis that the error variance of the dependent variable is equal across groups. a. Design: Intercept + YEARS + ENVIRON + YEARS * ENVIRON Tests of Between-Subjects Effects Dependent Variable: Job satisfaction Source Type III Sum of Squares df Mean Square F Sig. Corrected Model 205.778a 8 25.722 15.131 .000 Intercept 5467.022 1 5467.022 3215.895 .000 YEARS 18.311 2 9.156 5.386 .009 ENVIRON 12.311 2 6.156 3.621 .037 YEARS * ENVIRON 175.156 4 43.789 25.758 .000 Error 61.200 36 1.700 Total 5734.000 45 Corrected Total 266.978 44 a. R Squared = .771 (Adjusted R Squared = .720) Estimated Marginal Means 1. Length of employment Dependent Variable: Job satisfaction Length of employment Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound < 5 10.133 .337 9.451 10.816 5 - 10 11.333 .337 10.651 12.016 11 or more 11.600 .337 10.917 12.283 2. Nature of working environment Dependent Variable: Job satisfaction Nature of working environment Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound Highly structured 11.600 .337 10.917 12.283 Moderately structured 10.333 .337 9.651 11.016 Unstructured 11.133 .337 10.451 11.816 MGT223 MGT223 QUESTION 7 – SPSS OUTPUT (page 2 of 2) Length of employment * Nature of working environment Dependent Variable: Job satisfaction Length of employment Nature of working environment Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound < 5 Highly structured 13.600 .583 12.417 14.783 Moderately structured 9.600 .583 8.417 10.783 Unstructured 7.200 .583 6.017 8.383 5 - 10 Highly structured 11.800 .583 10.617 12.983 Moderately structured 10.800 .583 9.617 11.983 Unstructured 11.400 .583 10.217 12.583 11 or more Highly structured 9.400 .583 8.217 10.583 Moderately structured 10.600 .583 9.417 11.783 Unstructured 14.800 .583 13.617 15.983 Profile Plots MGT223 MGT223 QUESTION 8 – SPSS OUTPUT (page 1 of 3) Graph Correlations Delay Industry Public Quality Finished F in is h e d Q u a lit y P u b lic In d u s tr y D e la y Correlations 1 .287 -.180 -.264 -.624** . .073 .265 .100 .000 40 40 40 40 40 .287 1 -.137 .178 -.004 .073 . .398 .272 .979 40 40 40 40 40 -.180 -.137 1 .120 .062 .265 .398 . .462 .702 40 40 40 40 40 -.264 .178 .120 1 .036 .100 .272 .462 . .827 40 40 40 40 40 -.624** -.004 .062 .036 1 .000 .979 .702 .827 . 40 40 40 40 40 Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N Delay Industry Public Quality Finished Delay Industry Public Quality Finished Correlation is significant at the 0.01 level (2-tailed).**. MGT223 MGT223 QUESTION 8 – SPSS OUTPUT (page 2 of 3) Regression Variables Entered/Removedb Finished, Industry, Public, Quality a . Enter Model 1 Variables Entered Variables Removed Method All requested variables entered.a. Dependent Variable: Delayb. Model Summary .750a .562 .512 8.344 Model 1 R R Square Adjusted R Square Std. Error of the Estimate Predictors: (Constant), Finished, Industry, Public, Quality a. ANOVAb 3132.926 4 783.231 11.249 .000a 2436.974 35 69.628 5569.900 39 Regression Residual Total Model 1 Sum of Squares df Mean Square F Sig. Predictors: (Constant), Finished, Industry, Public, Qualitya. Dependent Variable: Delayb. Coefficientsa 88.784 4.519 19.647 .000 8.253 2.901 .328 2.845 .007 -1.759 3.231 -.062 -.545 .590 -2.307 .904 -.293 -2.552 .015 -7.680 1.414 -.609 -5.430 .000 (Constant) Industry Public Quality Finished Model 1 B Std. Error Unstandardized Coefficients Beta Standardized Coefficients t Sig. Dependent Variable: Delaya. MGT223 MGT223 QUESTION 8 – SPSS OUTPUT (page 3 of 3) Regression Variables Entered/Removedb Finished, Industry, Quality a . Enter Model 1 Variables Entered Variables Removed Method All requested variables entered.a. Dependent Variable: Delayb. Model Summary .748a .559 .522 8.262 Model 1 R R Square Adjusted R Square Std. Error of the Estimate Predictors: (Constant), Finished, Industry, Qualitya. ANOVAb 3112.282 3 1037.427 15.197 .000a 2457.618 36 68.267 5569.900 39 Regression Residual Total Model 1 Sum of Squares df Mean Square F Sig. Predictors: (Constant), Finished, Industry, Qualitya. Dependent Variable: Delayb. Coefficientsa 88.529 4.451 19.891 .000 8.509 2.835 .338 3.002 .005 -2.379 .886 -.302 -2.686 .011 -7.724 1.398 -.612 -5.525 .000 (Constant) Industry Quality Finished Model 1 B Std. Error Unstandardized Coefficients Beta Standardized Coefficients t Sig. Dependent Variable: Delaya.
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