Statistics and Difference

BIO cc3 SUMMATIVE date 2 Introduction The report analyses the result of a study on workers from brick and roofing cover industries conducted by the Health and Safety Laboratory (HSL). HSL put cumulus few criterias to the workers which creation that neither of the workers from the roofing covers and brick industries should have worked in both the industries and that they did non smoke. The criterias put across was an assurance to attain trusty results.The essence of the study lies in detecting any expiration in the health of the workers in these industries (as identified by mobile phone constipation) if any and alike to determine if any relationship exists amidst the distance of divine service and the enter health effect. The Null possible action (Ho) states that no difference in the median(a) amongst the division- change cells of the workers from the brick and roofing cover industries is observed. Null Hypothesis for the correlation study too states that there is no correlation among the health effects of the workers and the condemnation period they have worked in the industries.Nonetheless the Alternative Hypothesis (H1) states that the median dowery of ill-treatd cell of the workers in the brick industry is different when compared to the median lot of discredited cells of workers of both the exercises. H1 for the correlation study states that correlation exists between the term period the workers have worked in the industry and their health effects. Analysis volition be carried push through with the help of the following 5 s angstromles * Worker ID * hop on * Department * Length of service * Percent mature of cell footing The to a high place samples are independent within and also between each other.To keep an accurate analysis of the data, the normality, box plan and straight-line relationship and independence of the statistical analysis will be checked. The Null or Alternative Hypothesis will be accepted or jilted on t he al-Qaeda of a statistical analysis, which will be utilize to analyse the median percentage of traumad cells got from the brick and roofing tile operations. Table 1 Descriptive Statistics of brick and tile operation workers percentage dishonored cells Variable N N* suppose SE Mean St Dev. Minimum Q1 Median Q3 Maximum % Damaged cells of roofing tile operation 27 0 1. 337 0. 210 1. 090 0. cc 0. 600 1. 00 1. 500 4. 700 % Damaged cells of Brick operation 38 0 1. 532 0. 179 1. 106 0. 200 0. 536 1. 370 2. 189 4. 562 Table 1 gives a descriptive data of the workers of the respective industries. As seen in the table above the % of damaged cells of the workers in the brick industry is higher(prenominal)(prenominal) when compared with the tile operation workers. The median percentage of brick industry workers is 1. 370 which is higher as compared to the brick operation workers which is 1. 100. The inter-quartile range which being the difference between Q3 and Q1 is highe r for the brick operation compared to that of the tile. fancy 1Box piece displaying %damage of cell in workers from both tile and brick industries. The figure above shows that the percentage-damaged cell for tile operators is lower when compared with the brick operators indicating a difference in the designate and median. Figure 1 shows a difference in the health hazard of the tile and brick workers. There is render of skewness in the dissemination of brick operators whereas the tile distribution is symmetric, as the median line for the brick operators has shifted away from the centre.The % cell damage in workers of the tile operation is closely grouped apart from the 2 extremum outliers when compared to the % cell damage of the brick workers, which is quite wide. For the above box darn the deficiency for a further analysis is to be carried out as the possibility potbellynot either be accepted neither rejected since the box plot tho denotes statistical measures (mean, medi an, Q1, Q3, max & min values) which are not ample to prove the difference between the devil sites. Figure 2 Histogram of the Tile and Brick operation data The % of damaged cells of the brick operation is higher when compared to the tile operation.This is concluded from the histogram above which exhibits that the bar values which is the % damaged cells for brick operation is higher than the bar value of the tile operation. We have utilize a histogram, as it is one of the important tools for a data analysis. Figure 3The examen For Equal Variance. The values of the estimated equal variances show no difference in the % cell damage of the workers from the brick and tile operations-value obtained from the Levenes es range is 0. 200 which is also higher than 0. 05 implies that the meditation of difference cannot be rejected.The value of the F-Test is 0. 952 which being higher than 0. 05 shows also shows no signs that the unimportant guesswork (H0) should be rejected and also that there is no difference between %cell damage of workers from brick and tile operations. The obtained values from the try for equal variance point out to an abnormal distribution of data stating the acceptance of the null hypothesis. because no clear evidence of a difference in the median among the % damaged cells in the workers of both the operations. Figure 4Normal Distribution represent For Brick And Tile Operation.Figure 4 illustrates a normal distribution graph for tile and brick operations. The figure above shows that the %damaged cells of brick and tile operations are not uniformly distributed, as the points are not scattered about a straight line. There is evidence that the residuals followed a skewed distribution and it can also be seen that the above graph does not follow any trend or pattern. The is no convincing evidence to reject the null hypothesis (H0) as the P-Value is lower than 0. 05 in Fig4. From the above facts it may be concluded that the residuals do not foll ow a normal distribution.A MANN WHITNEY mental bear witness will be employ to statistically analyse the data as the %damaged cells of workers in the tile operation shows that the data is not unremarkably distributed since the P-Value is lower than 0. 05 and also that the plots on the graph so no street any precise trend. MANN WHITNEY TEST Results & CI Of Tile & Brick Manufacturing Operations Table 2illuminates the number of samples used in the Mann Whitney running game and the obtained median for data of brick and tile manufacturing operations Sample type recite of sample Median Tile 27 1. 100Brick 38 1. 370 omen estimate for ETA1-ETA2 is 0. 200 95. 0% CI for ETA1-ETA2 is (-0. 323, 0. 800) W = 1319. 0 Test of ETA1 = ETA2 vs. ETA1 not = ETA2 is significant at 0. 3905 The test is significant at 0. 3903 (adjusted for ties). The results shows a confidence interval of 95% between 0. 323 and 0. 800 in the %damaged cells of workers In the brick and tile operations. Contrariw ise the difference in the median is 0. 200(estimated), which factor that 0. 200%(approximately) more % of damaged cells in workers of the brick operations than those of the tile operations.A 100% authentic analysis cannot be proven as the confidence interval (CI) is only 95%, hence creating a need for more data in order to achieve a 100% certain analysis. An analyses of results obtained shows the P-value got from the Mann-Whitney test was 0. 3905. Since the P-value is higher than 0. 05 it indicated no evidence to reject the null hypothesis of no differences. thence it can be concluded that there is no convincing evidence of difference in the median between %damaged cells of workers in the 2 operations. ConclusionA use of several(a) graphs and descriptive statistics were used and inferred to make up ones mind if there were any differences in the health of the workers of the 2 operations. The Mann Whitney U test was considered to find the difference in the %-damaged cells of the tile and brick operation workers. A polish may be drawn from the these analyses that there is scarce evidence to mention that there is noteworthy difference in the % damaged cells in workers of tile and brick operations. Question 2 Table 3 Paired T-test and 95% CI to determine if the data of % damaged cells and duration of service of workers in two operations is paired. N Mean StDev SE Mean % Damaged cells 65 1. 451 1. 095 0. 136 aloofness of service (years 65 8. 995 7. 349 0. 912 Difference 65 -7. 544 6. 964 0. 864 95% CI for mean difference (-9. 270, -5. 819) T-Test of mean difference = 0 (Vs. not = 0) T-Value = -8. 73 P-Value = 0. 000 The table shows the T-test and the P-value got is >0. 05 stating no convincing evidence to reject null hypothesis of no differences. It may be concluded that the data is paired since the P-value is 0. 000. A scatter plot may also be used to test the relationship between the two samples.Figure5 A scatter plot screening the correlation betwe en the % of cells damaged with a regression line and the length of service in years. The predicted value for Regression is 17. 4%, which states the 17. 4% of the variability in the data is represented by the regression model. This cannot be used to get future values as the predictive value itself is rattling low. Pearsons correlation needs to be conducted since the above scatter plot shows a minor positive familiarity between the % damaged cells and the length of the service, but the damage of the cells in the future cannot be predicted.Pearsons correlation coefficient results Difference 65 -7. 544 6. 964 0. 864 95% CI for mean difference (-9. 270, -5. 819) T-Test of mean difference = 0 (vs. not = 0) T-Value = -8. 73 P-Value = 0. 000 Pearson correlation of length of service (years) and % damaged cells = 0. 417 P-Value = 0. 001. The association between the length of service and %damaged cells of the tile and brick operations cannot be accepted since the values from Pearsons Correl ation is 0. 417which is higher than 0. 400. Therefore a regression fitted line will be used to forecast the future data.The P-value is 0. 001 which being less than 0. 05 does not prove to be a convincing evidence to reject null hypothesis (H0) of no differences. Hence a conclusion may be drawn stating a difference in the length of services and the % damaged cells of workers from both the operations. Hence a regression fitted line plot will be used to predict future values. Further Analysis Figure6shows the data between the %damaged cells and the age of workers as well as the regression line. The scatter plot above shows that there is a moderate positive correlation between the age and the % damaged cells.Therefore a Pearsons correlation will be conducted. Pearson correlation of age (years) and % damaged cells = 0. 251 P-Value = 0. 044 The P value is 0. 044 which is less than 0. 05, this means that the null hypothesis must be rejected and the alternative hypothesis is accepted that t here is not sufficient evidence available to say that there is a correlation. Conclusion The data was analysed using descriptive statistics, various graphs, Pearsons correlation and regression fitted line plot to find association between the % damaged cell and length of service in tile and brick operations.The results concluded that there is no association between the % of damaged cells and their length of service. However there was a positive correlation which was observed between the % of damaged cells and age of workers in both operations. This suggested that it is the age which is the cause of damage and not the dust. The first test carried out, concluded that there is no authoritative difference between the health hazard of the worker at the tile and brick operation.The second test concluded that there is little relationship between the workers health and the length of their service. Since the R-sq value was only 17. 4%, the extent of damage cannot be predicted by the length o f employment. Overall conclusion It can be concluded that there is insignificant difference in the percentage damaged cells in the workers of tile and brick operations. It can also be concluded that age of workers and not the length of exposure to the dust in brick or tile operations increase % damaged cells of workers.