Business Decision-Making Issues - Assignment Example

Part Analysing the results of the Travel Newton & Oldingham survey Question 2: How long did it take you to travel to your departure station? Comment on the descriptive statistics for the length of time that passengers had to travel to the departure stations. How accurate are these statistics? Descriptive Statistics Variable N Minimum Maximum Mean SE SD time to departure station 250 2 35 16.68 0.32 5.03 Interpretation: The average time to departure station is 16.68 minutes with minimum of 2 minutes and maximum of 35 minutes. The standard deviation is 5.03 minutes. Previous research indicates that the average length of time passengers have to travel to start their train journeys is 20 minutes. Do your results match this result? Here we have to frame the test as follows: Null Hypothesis H0: There is no significant difference between the sample mean (average time to departure station 16.68) and the population mean (20 minutes). The population SD is not known, so we can utilize sample SD (standard deviation). The sample size is 250. Alternative Hypothesis H1: There is a significant difference between the sample mean and the population mean. The population SD is not known, so we can utilize sample SD (standard deviation). The sample size is 250. Level of Significance: 5% level or α=0.05. Test Statistic: Z0=(xbar – μ)/S/sqrt(n)= (xbar – μ)* sqrt(n)/S=(16.68-20)*sqrt(250)/5.03 =10.44 with probability 0. Hence we reject the null hypothesis and accept the alternative hypothesis and conclude that there is a significant difference between the sample mean (average time to departure station 16.68) and the population mean (20 minutes, survey already conducted). Conclusion: The average time to departure station is significantly lower than the estimated mean time to departure. Are there any differences in the average length of time that male and female passengers take to travel to the departure stations to start their train journeys? Group Statistics Variable gender N Mean SD SE(Mean) time to departure station male 142 16.66 5.16 0.43 female 108 16.70 4.87 0.47 Variable F Sig. T df Sig. (2-tailed) Mean Difference Std. Error Difference time to departure station 0.057 0.812 -.065 248 0.948 -0.0417 0.64 Null Hypothesis H0: There is no significant difference in the time to departure station between males and females. Alternative Hypothesis H1: There is a significant difference in the time to departure station between males and females. Level of Significance: 5% level or α=0.05. Test Statistic: t0=(x1bar – x2bar)/SE(x1bar – x2bar) SE(x1bar – x2bar)=sqrt[((n1-1)S12+(n2-1)S22))/(n1+n2-2)] where S12 is first sample variance and S22 is second sample variance. Based on the above table results, the probability of significance is 0.948 (>0.05), we accept our null hypothesis and conclude that the mean time to departure station between males and females is not significantly different. They are on par. Conclusion: Males and females have the same mean time to departure station. Question 4: How many times a month do you make this journey by train? Comment on the descriptive statistics for the number of times a month passengers make their journeys. How accurate are these results? Descriptive Statistics Variable n Minimum Maximum Mean SE(Mean) SD times journey made each month 250 1 24 12.06 .52 8.22 The mean number of times a month passengers make their journeys is 12 times with minimum 1 travel and maximum 24 travels and the standard deviation is 8.22. Previous research indicates that the average number of times that passengers make journeys on this train service is 14 times a month. Do your results match this result? Null Hypothesis H0: There is no significant difference between the sample mean (average number of times 12.06) and the population mean (14 times). The population SD is not known, so we can utilize sample SD (standard deviation). The sample size is 250. Alternative Hypothesis H1: There is a significant difference between the sample mean and the population mean. The population SD is not known, so we can utilize sample SD (standard deviation). The sample size is 250. Level of Significance: 5% level or α=0.05. Test Statistic: Z0=(xbar – μ)/S/sqrt(n)= (xbar – μ)* sqrt(n)/S=(12.06-14)*sqrt(250)/8.22 =3.73 with probability 0.000096 Hence we reject the null hypothesis and accept the alternative hypothesis and conclude that there is a significant difference between the sample mean (average no. of times a month travelled i.e.12.06) and the population mean (14 times). Conclusion:The average no. of times a month travelled is significantly lower than the estimated average. Are there any differences in the average number of times that male and female passengers make the journeys? Group Statistics Variable Gender N Mean SD SE(Mean) times journey made each month male 142 11.87 8.15 0.68 female 108 12.32 8.35 0.80 Independent Samples Test Variable F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference times journey made each month 0.420 0.517 -0.436 248 0.664 -0.46 1.05 Null Hypothesis H0: There is no significant difference in the times journey made each month between males and females. Alternative Hypothesis H1: There is a significant difference in the times journey made each month between males and females. Level of Significance: 5% level or α=0.05. Test Statistic: t0=(x1bar – x2bar)/SE(x1bar – x2bar) SE(x1bar – x2bar)=sqrt[((n1-1)S12+(n2-1)S22))/(n1+n2-2)] where S12 is first sample variance and S22 is second sample variance. Based on the above table results, the probability of significance is 0.517 (>0.05), we accept our null hypothesis and conclude that the mean times journey made each month between males and females is not significantly different. They are on par. Conclusion: The mean times males make journey is the same as that of mean time females make the journey. Question 5: What kind of ticket did you use? Type of ticket used S.No. Frequency Percent 1 6 2.4 2 75 30.0 3 15 6.0 4 95 38.0 5 30 12.0 6 29 11.6 Total 250 100.0 Interpretation: It is interpreted that 38% of the passengers use type 4 tickets, followed by 30% of the passengers who use type 2 ticket and 12% of the passengers who use type 5 tickets. Question 6. What is the main reason for your journey? Do the results of these two questions indicate any pattern? Can we say that certain types of tickets are used for particular reasons for travelling? Main reason for journey Purpose of journey Frequency Percent Work 101 40.4 Education 29 11.6 Shopping 43 17.2 Personal Business 24 9.6 Visiting Friends or Relatives 20 8.0 Hospital or Doctors 13 5.2 Social or Leisure 14 5.6 Others 6 2.4 Total 250 100.0 Interpretation: From the above table it is observed that 40.4% of the respondents travel to work, 17.2% of the respondents travel for shopping and 11.6% of the respondents travel for education. Conclusion: Majority of the respondents’ purpose of journey is for work. Type of ticket used * main reason for journey Cross tabulation main reason for journey Total Work education Shopping personal business visiting friends or relatives hospital or doctors social or leisure type of ticket used 1 5 1 6 2 4 33 13 7 6 8 71 3 6 4 3 2 15 4 95 95 5 2 4 2 9 5 6 28 6 29 29 Total 101 29 43 24 20 13 14 244 Interpretation: It is observed that 95 respondents use type 4 ticket for work, 33 respondents use type 2 ticket for ticketing and 29 respondents use type 6 ticket for education. Question 7: Thinking about the station you travelled FROM, please rate how satisfied you are with the following aspects [facilities, rail services, accessibility, public transport information, personal safety] For each of the factors, facilities, rail services etc., is there any difference in the responses based on which station the passengers had travelled from? Station travelled from * rating on facilities Cross tabulation rating on facilities Total very satisfied fairly satisfied neither fairly dissatisfied very dissatisfied station travelled from 1 23 20 13 5 6 67 2 17 23 9 7 8 64 3 21 12 6 4 6 49 4 18 23 14 9 6 70 Total 79 78 42 25 26 250 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 8.276 12 .763 Conclusion: From the above tables, it is found that there is no association between station travelled from and the rating on facilities. Station travelled from * rating on rail services rating on rail services Total 1 2 3 4 5 station travelled from 1 23 20 12 7 5 67 2 21 24 11 5 3 64 3 19 11 4 6 9 49 4 18 24 10 9 9 70 Total 81 79 37 27 26 250 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 12.911 12 0.376 Conclusion: From the above tables, it is found that there is no association between station travelled from and the rating on rail services. station travelled from * ratings on accessibility ratings on accessibility Total 1 2 3 4 5 station travelled from 1 27 16 9 7 8 67 2 14 10 22 11 7 64 3 15 9 15 6 4 49 4 14 18 18 9 11 70 Total 70 53 64 33 30 250 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 17.162(a) 12 0.144 Conclusion: From the above tables, it is found that there is no association between station travelled from and the rating on accessibility. station travelled from * ratings on transport information ratings on transport information Total very satisfied fairly satisfied neither fairly dissatisfied very dissatisfied station travelled from 1 20 23 15 4 5 67 2 21 20 10 9 4 64 3 15 10 11 5 8 49 4 21 21 11 8 9 70 Total 77 74 47 26 26 250 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 9.429 12 0.666 Conclusion: From the above tables, it is found that there is no association between station travelled from and the rating on transport information. station travelled from * ratings on personal safety ratings on personal safety Total 1 2 3 4 5 station travelled from 1 7 15 14 19 12 67 2 21 18 11 5 9 64 3 14 10 10 8 7 49 4 17 17 15 10 11 70 Total 59 60 50 42 39 250 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 17.943(a) 12 0.117 Conclusion: From the above tables, it is found that there is no association between station travelled from and the rating on personal safety. Is there any difference in the responses regarding personal safety between the male and female passengers? ratings on personal safety * gender Cross tabulation gender Total Male female ratings on personal safety 1 34 25 59 2 34 26 60 3 26 24 50 4 26 16 42 5 22 17 39 Total 142 108 250 Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square .935 4 .920 Interpretation: Since the chi square value of 0.93 with probability 0.92 (>0.05), is not significant, we accept our null hypothesis and conclude that there is no association between gender and ratings on personal safety. Is there any difference in the proportion of females who are “very dissatisfied” with their personal safety compared to male passengers? The proportion of males who are very dissatisfied are 22/142 and that of females is 17/108. So from proportions test between males dissatisfied and females dissatisfied, the Z value is observed to be 0.0746 with probability of significance 0.47 (>0.05). Hence it is concluded that there is no significant difference between the proportion of males and females in the highly dissatisfaction on ratings on personal safety. Part 2 Modelling Record Sales Model Summary Model R R Square Adjusted R Square Std. Error of the Estimate 1 0.815 0.665 .660 47.09 ANOVA Sum of Squares df Mean Square F Sig. Regression 861377.418 3 287125.806 129.498 0.000 Residual 434574.582 196 2217.217 Total 1295952.000 199 Predictors: (Constant), Attractiveness of Band, Advertising Budget (thousands of pounds), No. of plays on Radio 1 per week Dependent Variable: Record Sales (thousands) Coefficients Unstandardized Coefficients B Std. Error Standardized Coefficients Beta t Sig. (Constant) -26.613 17.350 -1.534 0.127 Advertising Budget (thousands of pounds) 0.08488 0.007 0.511 12.261 0.000 No. of plays on Radio 1 per week 3.367 0.278 0.512 12.123 0.000 Attractiveness of Band 11.086 2.438 0.192 4.548 0.000 a Dependent Variable: Record Sales (thousands) Interpretation: The regression model is a good fit since the multiple correlation coefficient is 0.665 and is highly significant, (probability 0.000) and the multiple R2 value is 0.665 which indicates that 66.5% of the dependent variable is explained by the independent variables. From the above regression model (linear), the least square fit of expected sales (y, dependent variable) against the independent variables advertising budget (x1), no. of plays on radio 1 per week (x2) and attractiveness of band (x3) is as follows: Y=0.08488*x1+3.367*x2+11.086x3–26.613 The model is a good fit since the standard error for the independent variables are lower, except the independent term because for all the other variables the probability of significance is 0.000 and for independent term the probability of significance is 0.127. Conclusion: The linear regression model is a good model since the prediction of sales based on the other independent variables is highly significant since the scatter plot is showing a good closeness of observed and expected values in the positive direction (XY plane). Read More