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程序代写案例-MGT223 1
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MGT223 1 Material Provided: Statistical Tables, Appendix of SPSS output, Formula sheet > MANAGEMENT SCHOOL Spring Semester 2017-18 BUSINESS STATISTICS 2 Hours Where appropriate, candidates are expected to provide a confidence interval, undertake a statistical test, specify the assumptions and interpret the result. Answer ALL questions. The marks allocated to each question and, where appropriate, sub-parts are indicated. MGT223 1. The manager of a hotel claims that the average guest bill for a weekend is only £250. A local journalist, however, claims that prices have increased and that average bills are greater than this. A sample of 26 bills from the previous weekend gave a mean of £278 with a standard deviation of £22.50. Does this disprove the manager’s claim? (12) Null -H0: Alternative - H1: (is the manager under stating the average cost?) n =26 ̅ = 278 s = 22.50 n < 30 – use t-distribution. Assumptions: random sample, normal distribution 95% CI: ̅ ± −1,0.025 × √ 278 ± 25,0.025 × 22.5 √26 278 ± 2.0595 × 4.4126 278 ± 9.08 The CI for the average bill ranges £269 to £287 – the data are not consistent with the null hypothesis of the average being £260 – how strong is the evidence? p-value = P(̅ > 278|H0) ( ≥ 278 − 250 22.5 √26 ⁄ ) ( ≥ 6.345) From tables: P(t25 > 3.7251) = 0.0005. Thus p-value < 0.0005 (0.05%) Very strong evidence against the null hypothesis. Based on the sample evidence there is very strong evidence against the manager’s claim – the best estimate of the average bill is £269 to £287. However we may have some reservations over whether the sample is truly random – bills from previous weekend. MGT223 3 2. A health food cafe operator reads that a sample survey by the National Restaurant Association shows that 45% of adults are committed to eating nutritious food when eating away from home. To help plan her menu she decides to conduct a survey among her own customers. She believes that the nutritional value of food is important to a greater proportion than stated by the survey. From an initial sample of 16 customers, 12 agreed with the statement: “The nutritional value of food is important to me when I eat out”. Based on this evidence she concluded that her customers definitely place more importance on the nutritional value of their food than the national average. Is she correct? (12) Null -H0: Alternative - H1: (her customers more likely to focus on nutritional value) n =16 12 agree The sample size is small so use Binomial directly. 95% CI for true proportion (from charts): 0.465 to 0.93. Very wide but does not include the hypothesised value of 45% so data not consistent with null. p-value = P(number agreeing ≥ 12| H0: P(number agreeing ≥ 12| Binomial n=16, 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 45%. The best estimate is 46.5% to 93%. 3. A researcher was interested in understanding the stress levels of board members in large corporations during supplier negotiations. In one study she took a group of nine directors and measured their anxiety (out of 15) during a normal supplier meeting and then again in another meeting in which she had paid a bogus supplier to be deliberately uncooperative. The data was analysed in IBM SPSS Statistics using a variety of tests and the results are given in the Appendix a) Provide a full interpretation of these analyses, indicating the hypotheses that you are testing and the assumptions underpinning the tests. (22) Output from Explore: Average anxiety for normal situation appears to be lower (mean = 6.87, median = 7 compared with 9 and 8 respectively for uncooperative situation) Appears to be more variability in response to uncooperative supplier (sd = 1.208 vs 3.606, range = 3 vs 10) These differences are reflected in the boxplots and we can also see a lack of symmetry in the Normal supplier responses (2) Interpretation of tests of normality and plots below: Two sample t-test: Two sample t-test assumes normal distributions, data at least interval and equal variances. Interval data – 15 point scale – assume interval Check normality via the Explore output: Values for skewness and kurtosis provide no evidence against normality Normal: -0.451 /.717 = -0.629, -1.237/1.400 = -0.884 – both in range -2 to 2 Uncooperative: 0.720 /.717 = 1.004, -0.66/1.400 = -0.471 – both in range -2 to 2 Shapiro-Wilk tests provides some evidence against normality for the Normal supplier observations (p = 0.048) but no evidence against for Uncooperative (p = 0.180) Limited number of points in the Normal probability plots – they appear to be fairly linear. Lack of symmetry in boxplots, particularly for normal supplier observations. So limited data (n=9) but some suggestion of non normality – proceed with parametric tests with caution. Check equal variances with Levene’s test H0 H122 p = 0.012 – good evidence against equal variances – we cannot assume this so look at second row of table MGT223 5 (5) Main test H0: H1: 2 (higher stress levels with uncooperative suppliers but might argue for two-sided – in which case p-values should be two-sided) (1) 95% CI for difference (-4.97 to 0.700). This includes 0 but only just. The data are consistent with the null hypothesis but stress levels with uncooperative suppliers might be higher by almost 5 units on average. Looking at the p-value (one-sided): p < 0.062. Slight evidence against null hypothesis. There is some (weak) evidence against the null hypothesis. Stress levels might be almost 5 units higher on average but the difference could be very small. Effect size = eta-squared = 2 2+1+2−2 = 1.6832 1.6832+9+9−2 = 0.154 – a large effect. (2) Mann Whitney Test Assumes data can be ranked i.e. ordinal (appropriate here). H0: samples drawn from sample population distribution H1: samples drawn from different population distributions. p-value (one-sided) = 0.111 No evidence against the null hypothesis. (2) Paired tests Output from Explore: Average difference in anxiety levels is negative: mean = -2.133, median = -1.00 Difference varies from -7 through to 1 with sd = 2.74 (1) Interpretation of test of normality and plots below: Paired t-test: Paired t-test assumes differences normally distributed, data at least interval and data paired. Interval data – 15 point scale – assume interval Check normality via the Explore output: Values for skewness and kurtosis provide no evidence against normality -0.961 /.717 = -1.34 , -.205/1.400= -0.143 both in range -2 to 2 Shapiro-Wilk tests provides no evidence against normality (p = 0.191) Limited number of points in the Normal probability plots – they appear to be fairly linear. Slight lack of symmetry in boxplots, but no real suggestion of non normality – proceed with parametric test. (2) MGT223 7 Correlation indicates efficacy of pairing. There is good evidence of positive correlation between the two sets of scores (r = 0.798, p=0.01) – Stress levels do vary by director so pairing the data is appropriate and should lead to a more powerful test. (2) H0d H1d (1) The 95% CI ford Ranges from (-4.24 to -0.027). The data are not consistent with the null hypothesis but the difference could be very small (0.027). Looking at the p-value =0.024, there is evidence against the null hypothesis. There is evidence that stress levels are higher with uncooperative suppliers (could be a very small difference or could be over 4 units on average). Effect size = eta-squared = 2 2+−1 = 2.3352 2.3352+9−1 = 0.405 – a large effect. (2) Wilcoxon Signed Ranks Test. H0: samples drawn from sample population distribution H1: samples drawn from different population distributions (higher scores for uncooperative suppliers). Data need to be at least interval as we are calculating differences between scores. Also need to be paired (as here). 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 is a difference (4.71 vs 3) and there are more negative ranks (7) than positive ones (1) p-value = 0.020 – good evidence against null hypothesis. Effect size r = 2.111/sqrt(18) = 0.498. A large effect. More negatives than positives and larger negative differences. There appears to have been an increase in stress levels with uncooperative suppliers. (2) b) Which of these tests is most appropriate? Explain your answer. The data are paired (description, backed up by r=0.798) so one of the paired tests would be appropriate. Incorporating information on pairing leads to a more powerful test as we can eliminate differences due to individuals and focus on the impact of the uncooperative supplier. (2) Assumptions underpinning paired t-test appear to hold so this might be preferable as it provides an estimate of the difference. Alternatively might argue that, given the small data set, the fewer the assumptions the better – so non parametric test preferable. (2) c) What conclusions can you draw about the stress levels of directors in supplier negotiations? There is good evidence (p = 0.024, effect size = 0.405 – a large effect size) that directors’ stress levels are higher with uncooperative suppliers. The average difference could be close to 0 but it could be almost 5 units on average. (3) There is also evidence that stress levels vary with directors – some tend to have higher levels than others. (1) 4. 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 and the form of the results. a) A large employer is considering changing work schedules for its employees. In order to assess support for the schemes available, 100 employees were asked to indicate whether or not they were in support of a four 9-hour day working week. How would you assess the popularity of this scheme? Confidence Interval for the population proportion: n = 100 (≥20). As long as the sample proportion is not too extreme (np and n(1- p) ≥ 5) use: This will give a range of plausible values for the true proportion in favour. (2) b) A subscriber survey in a Business magazine asked respondents to indicate the type of ticket (First class, Business, Economy) and type of flight (Domestic, International) they purchased most frequently for business purposes. They were interested in whether a link existed between type of flight and type of ticket. Chi-square test of association. Null -H0: No association between ticket type played and flight type. Alternative - H1: Some link between ticket type played and flight type. Crosstabulation of results – look at percentage breakdowns. (2) n pp p )1( 96.1 MGT223 9 c) The standard deviation of earnings per share is available for samples of 10 companies in the airline industry and 12 in the automobile industry. How would you compare the variability of earnings per share between the two industries? F-test of equality of variances: Null -H0: Alternative - H1: ≠ − = 2 × ( ≥ 2 2 ) And look up in F tables (2) d) The manager of a water-filtration systems service company would like to estimate more accurately the likely service time for each maintenance request. Repair time is believed to be related to two factors, the number of months since the last maintenance service and the type of repair problem (mechanical or electrical). Data for a sample of 30 repairs is available. Regression analysis. Dependent variable – service time with Explanatory variables of number of months and a dummy (0/1)variable representing type of problem (2) e) An investigation was undertaken into the performance of two computerised language translators. Translations were made with both systems for three different languages. Three standard texts were translated for each language/translator combination and the length of time taken to translate each text was recorded. How would you analyse this data? Two way ANOVA with interaction. Test for translator/language interaction and then the main effects of language and translator. (2) 5. A study provided data on variables that may be related to the number of weeks a manufacturing worker has been jobless due to layoff. The dependent variable in the study (WEEKS) was defined as the number of weeks a worker has been jobless due to a layoff. The following independent variables were used in the study: AGE: The age of the worker EDUC: The number of years of formal education HEAD: A dummy variable: 1 if the head of the household, 0 otherwise TENURE: The number of years in last job OCCUP: 1 if management occupation, 2 if sales occupation, 3 other occupation The data set was analysed using the IBM SPSS Statistics and the results are presented in the Appendix. a) Before applying regression techniques, the statistician created two new 0/1 variables: MANAGEMENT: 1 if employee in Management occupation; 0 otherwise. SALES: 1 if employee in Sales occupation; 0 otherwise. Explain why this was necessary. (4) The variable OCCUP is a nominal variable – the numbers are simply labels. In order to include the information on occupation into the regression equation the information contained in this nominal variable must be converted into two 0/1 variables, representing two of the three occupation categories. (2) The statistician has chosen to represent the categories Management and Sales in this form. These variables are referred to as dummy variables and can be included in the regression model. (2) MGT223 11 b) Fully analyse the output. (16) The scatter plot shows evidence a positive correlation between WEEKS and AGE. There is also some evidence that being the head of the household leads to fewer jobless weeks and that weeks jobless may be influenced by occupation with Other occupations (OCCUP = 3) tending to have higher values. There is no clear relationship between weeks jobless and education or years in previous job. The correlations observed in the scatterplots are confirmed in the table of correlations. There is a moderate positive correlation between WEEKS and AGE (r=0.564, p<0.0005) but no evidence of correlations among the other explanatory variables. (3) Regression Three regression analyses are presented. The first has an R Square value of 0.661 and adjusted R Square of 0.623 indicating good explanatory power but years of formal education is not significant (p = 0.724). (2) Dropping EDUC, the second has an R Square value of 0.661 and adjusted R Square of 0.629, a slight improvement on the first model, indicating good explanatory power but years in previous job is not significant (p = 0.136). (2) Dropping TENURE, the third has an R Square value of 0.646 and adjusted R Square of 0.620, a slight drop in explanatory power compared with the previous models but now all explanatory variables are significant (AGE, p < 0.0005; HEAD, p = 0.029; MANAGEMENT, p=0.002 and SALES, p < 0.0005). (2) Equation: WEEKS = 3.14 + 1.24*AGE – 7.10*HEAD – 25.18*MANAGEMENT – 20.42*SALES (1) Older workers tend to have increased weeks jobless but being head of household, in a Management or Sales occupation are all associated with fewer weeks jobless (reductions of 7, 25 and 20 weeks respectively) (2) The Beta values give some indication of the relative importance of each of the explanatory variables; they indicate the number of standard deviations WEEKS will increase for a one sd rise in each explanatory variable. The value of Beta for AGE is 0.573 compared with -.559 for SALES, -.266 for MANAGEMENT and - 0.188 for HEAD. (1) The residual plots allow us to assess the assumptions of linearity, normality and homoscedasticity (constant variance) which underpin the regression analysis. They also allow us to identify any outliers. Looking at the histogram and normal probability plot, there is no clear evidence of non normality. An inspection of the scatterplot provides no indication of heteroscedasticity or non linearity. There are no obvious outliers. (3) c) Write a brief report, suitable for a non-statistician, summarising your findings. (4) We have used regression analysis to investigate the relationship between weeks jobless following layoff (WEEKS) and age (AGE), length of time in previous job (TENURE), years of formal education (EDUC), whether or not the individual is identified as head of household (HEAD), and occupation (MANAGEMENT, SALES or OTHER). (1) The following equation was derived: WEEKS = 3.14 + 1.24*AGE – 7.10*HEAD – 25.18*MANAGEMENT – 20.42*SALES TENURE and EDUC offered no additional explanatory power so were not included. (1) There is strong evidence that age is lined to increased weeks jobless (1.24 weeks for each additional year) and being in a Management or Sales occupation is linked to fewer weeks jobless (25 and 20 weeks respectively).. There is good evidence that those identified as head of household experience fewer weeks jobless (7 weeks) (1) The predictive power of this equation is reasonably good - around 65% of the variability in weeks jobless can be explained by the equation. (1) MGT223 13 6. A manufacturing company designed an experiment to determine whether the number of defective parts produced by two machines differed and if the number of defective parts produced also depended on whether the raw material needed by each machine was loaded manually or by an automatic feed system. The data set was analysed using the IBM SPSS Statistics and the results are presented in the Appendix. a) Fully analyse the output, stating any assumptions that you must make. (8) Assume normal populations and homogeneity of variance. Normality is difficult to assess but there is no evidence against equal variances (p=0.423). (2) Interaction H0: no interaction between loading method and machine H1: some interaction between loading method and machine p-value = 0.099, partial eta-squared = 0.130 – weak evidence of an interaction, large effect size (1) Main effects H0: no additional machine effect H1: some additional machine effect p-value <0.0005 – very strong evidence against null – the machines do differ in their defect rates. The effect size is large (partial eta-squared = 0.616). (1) H0: no loading method effect H1: some loading method effect p-value = 0.041 – some evidence against null – loading method does seem to have an impact on defect rates. The effect size is large (partial eta-squared = 0.193). (1) So main conclusions are that there is evidence that the number of defects varies according to machine used (very strong evidence) and loading method (some evidence) and weak evidence of an interaction effect between machine and loading method. Each factor has a large effect size. (1) The descriptive statistics give mean defect numbers for individual factors and combinations. It appears that defect rates are highest for Machine 1 and that Automatic loading results in higher defect rates for both machines. (1) The profile plot reinforces this and it is possible to see that the combination of Machine 2 with manual loading is particularly effective, leading to the lowest mean overall. (1) b) Write a brief report, suitable for a non-statistician, summarising your findings. The data were analysed using two way analysis of variance. There was very strong evidence of a difference between the two machines. There was good evidence of a loading effect and weak evidence of an interaction between loading method and machine. All effect sizes were large. (2) Machine 2 resulted in fewer defectives for both loading methods and the manual loading method resulted in fewer defectives for both machines. The combination of Machine 2 with manual loading was particularly effective in reducing defects. (2) (4) End of Question Pape MGT223 MGT223
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