Problem-Solving Statistics Analysis - Assignment Example

Problem Solving with Statistics Part a) The graph below shows the relationship between GDP and logistics bill for eleven European countries. (b) The graph indicates that there is a relationship between GDP and Logistics Bill. The countries with lower GDP have lower logistics bills and those with higher GDP have higher logistics bills. The data in the table below also reveal some interesting statistics. (c) The table below shows the regression statistics which corresponds to the 95% confidence interval for the regression coefficient. SUMMARY OUTPUT Regression Statistics Multiple R 0.994761482 R Square 0.989550405 Adjusted R Square 0.988389339 Standard Error 5801.482154 Observations 11 ANOVA   df SS MS F Significance F Regression 1 28685267062 2.87E+10 852.2774 3.17E-10 Residual 9 302914756.7 33657195 Total 10 28988181818         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 90.0% Upper 90.0% Intercept -727.555302 2557.155538 -0.28452 0.782452 -6512.24 5057.132 -5415.11 3960 GDP 0.123935096 0.004245255 29.19379 3.17E-10 0.114332 0.133539 0.116153 0.131717 The information in the table indicates that there is a 99% correlation between both variables. It means that 99% of the variation in the logistics bill (y) is explained by GDP (x). This is suggestive of a strong positive correlation between logistics bill and GDP for the 11 European countries. The regression equation is: y = 0.1239x - 727.55 This equation defines the relationship between two variables – in this case GDP and logistics bill The formula indicates that for every $1 change in GDP the logistic bills changes by $0.124. The 95% confidence interval for this relationship suggests that the logistics bill will lie between 11% and 13% of GDP. That is we can say with 95% confidence that the bill will be between 11% and 13% of GDP (d) The logistics bill for a country with a GDP of $600,000mn is found by substituting the GDP which is x into the following formula: y = 0.1239X – 727.55 y = 0.1239 x $600,000 – 727.55 y = $74,340 – $727.55 y = $73,612 GDP Logistics Bill 161,800 19,000 104,100 14,000 956,900 110,000 1,189,100 160,000 53,400 7,000 33,900 5,000 856,800 102,000 223,600 27,000 45,300 6,000 375,700 45,000 832,600 96,000 600000 73612 SUMMARY OUTPUT Regression Statistics Multiple R 0.994826 R Square 0.989679 Adjusted R Square 0.988647 Standard Error 5503.773 Observations 12 ANOVA   df SS MS F Significance F Regression 1 2.9E+10 2.9E+10 958.939 2.89E-11 Residual 10 3.03E+08 30291518 Total 11 2.94E+10         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept -728.589 2409.933 -0.30233 0.768597 -6098.25 4641.076 -6098.25 4641.076 GDP 0.123933 0.004002 30.96674 2.89E-11 0.115016 0.132851 0.115016 0.132851 The logistics bill for a country with a GDP bill of $600,000mn is $73,612mn. (e) The 95% confidence interval for the prediction in (d) is found from the following table. The information in the table was generated from Microsoft Excel using the regression tool. It indicates that the confidence interval lies between 0.115 and 0.133. This means that we can state with 95% confidence that the interval from 0.115 to 0.132 contains the percentage of the logistics bill relating to GDP of $600,000mn. (f) The data suggests that the line does not pass through the origin because the intercept is a negative 727. This suggests that when GDP is equal to $0 the logistics bill is -$727. Part 2 Section (a) (i) The number of machines working at any time based on a sample of 30 activities is shown in the following table. Number of machines working out of a total of 34 machines X 22 20 16 26 24 19 27 21 20 31 22 27 29 25 28 30 22 23 28 32 20 26 19 20 27 24 16 27 22 21 714   ƩX = 714 n=30 Ẋ =ƩX/n = 714/30 =23.8 = 24 On average 24 machines are in use at any one time. The average utilisation is calculated as follows: (Average number of machines in use/Number of machines available) x 100% = 24/34 x100% = 70% (ii) A model that can be used to represent the activity of the machines is the exponential distribution. (iii) The lowest number of machines likely to be working at any one time is 16. (iv) The maximum number of machines likely to be working at any time is 32 … Section b A comparison of the results of the two methods used to test pollution levels were carried out. The table below shows the paired t-Test results for the nine days. t-Test: Paired Two Sample for Means   Firms Method EHAs Method Mean 34.55555556 39 Variance 96.77777778 131.75 Observations 9 9 Pearson Correlation 0.781540664 Hypothesized Mean Difference 0 df 8 t Stat -1.848506995 P(T10) = P(11) + P(12) . The probabilities for x = 11 and x = 12 were found using the binomial distribution function in Microsoft Excel. The results are as follows: Probability of x > 10 Probability of x = 11: n 12 r 11 p 0.9 Cumulative FALSE Result: P(11) = 0.377 Probability of x = 12: n 12 r 12 p 0.9 Cumulative FALSE Result: P(12) = 0.282 P(x > 10) = 0.659 (d) The mean (μ) and standard deviation (√npq) of a binomial distribution is computed as follows: μ = np = 12*.9 = 10.8 √npq = √10.8(1 - 0.9) = √10.8*0.1 = 1.08 The mean indicates on average how many of every 12 firms will have open order shipping file database while the standard deviation indicates that the distribution is within 1.08 standard deviation from the mean. Section (ii) (a) An appropriate distribution for modelling this data is a Poisson distribution. The situation meets the requirements for a Poisson distribution because of the following. The number of time the event occur is known The number of times the event could have occurred is not known and so the event is rare Observations are made over short time intervals There is an occurrence rate for the number of breakdowns per shift The population mean is equal to the population variance The number of breakdowns vary randomly about the average The formula for computing Poisson probabilities is: P(x) = *x/(x!)(e*) or P(x) = (e-*x)/x! Where x is the number of occurrences  is the rate of occurrences e is a constant with value 2.71828 (b) The probability of exactly two (2) breakdowns (P(2)) during the night shift is calculated in the table below. Probability x = 2 x 2 Mean 1.6 Cumulative FALSE Result: P(2) 0.258     The result in the table has been calculated with the use of Microsoft Excel. The probability of exactly two breakdowns in the night shift is 0.258. P(2) = 0.258 (b) The probability of less than three (3) breakdowns during the afternoon shift is calculated in the table below with the use of Microsoft Excel. Probability x < 3 probability of x = 0   x 0 Lambda() 1.6 Cumulative FALSE Result: P(0) 0.202     Probability x = 1   x 1 Lambda ( 1.6 Cumulative FALSE Result: P(1) 0.323     Probability x = 2   x 2 Lambda ( 1.6 Cumulative FALSE Result: P(2) 0.258 P( x < 3) = 0.783 The table shows that the probability of x < 3 (combined probabilities of x = 0, x = 1 and x = 2) is 0.202 + 0.323 + 0.258 = 0.783. P(x < 3) = 0.783 (d) The probability of no breakdowns during three consecutive shifts is calculated as follows. Probability of x = 0 for 3 consecutive shifts x 0 Lambda 1.6 Cumulative FALSE Result 0.202 Number of 8 hour shifts in a day 3 Probability (3*0.202) 0.606 There are three (3) 8 hour shifts in a day and the occurrence rate of 1.6 per 8 hour shift is the same. The probability of having no breakdowns in all three shifts in one day was found by multiplying the result (0.202) by 3 which yielded 0.606 as shown in the table above. Section (iii) (a) An appropriate model for describing the data is a hyper geometric distribution. The reason is that the probability of success varies from trial to trial does not remain the same as in a binomial distribution. Additionally, the batch size is large and it involves sampling without replacement. The formula for this model is: P(r) = (MCr * N - MCn – r)/NCn Where r is the number of defectives in a sample N is the batch size M s the number of defective items in the batch n is the number in the sample. (b) If the previous 50 batches had an average defective rate of 9 items per batch then the quality of the supplier would appear to be deteriorating since the calculated mean is 10.4 as shown below. Number of defects Frequency (f) FX 4 1 4 5 2 10 6 1 6 7 3 21 8 6 48 9 5 45 10 7 70 11 5 55 12 8 96 13 5 65 14 3 42 15 4 60 114 50 522 Mean = Ʃfx/Ʃf = 522/50 = 10.44 If the difference between the mean of the previous set of 50 batches and the last set of 50 batches is not statistically different then the quality has remained the same. The test which is used to determine statistical significance is the paired t-Test. The test was chosen because it compares two sets of 50 batches containing 100 items in each batch. The number of defective items in each of the previous fifty batches would be paired with each of fifty batches in the last fifty batches. The assumption is that the variances are different. …. Read More