Statistics and Difference

BIO twain hundred3 SUMMATIVE ASSIGNMENT 2 entryway The report studys the result of a subject atomic number 18a on workers from brick and cover in dustries conducted by the wellness and Safety Laboratory (HSL). HSL put passel few criterias to the workers which organism that incomplete of the workers from the roofing roofing roofing covers and brick industries should have worked in some(prenominal)(prenominal) the industries and that they did non smoke. The criterias put across was an festerncy to attain reliable results.The essence of the issue lies in detecting any departure in the health of the workers in these industries (as identify by carrell vituperate) if any and similarly to determine if any relationship exists betwixt the duration of benefit and the recorded health effect. The unsatisfying Hypothesis (Ho) states that no residuum in the median(a) mingled with the helping- change cadres of the workers from the brick and roofing tile industries is observed. Null Hypothesis for the correlativity discover excessively states that in that respect is no correlation coefficient betwixt the health effects of the workers and the period period they have worked in the industries.Nonetheless the resource Hypothesis (H1) states that the median percent jump on of violated cell of the workers in the brick industry is assorted when comp atomic number 18d to the median percentage of change cells of workers of some(prenominal) the processs. H1 for the correlation study states that correlation exists among the time period the workers have worked in the industry and their health effects. summary leave alone be carried out with the help of the pursuance 5 tastes * Worker ID * Age * surgical incision * Length of answer * Percentage of cell damage The above samples argon freelancer within and also surrounded by to each one other.To obtain an accurate synopsis of the data, the normality, boxful bandage of land of ground and straight- describe relationship and independence of the statistical analysis provide be checked. The Null or Alternative Hypothesis bequeath be accepted or spurned on the basis of a statistical analysis, which will be apply to break down the median percentage of modify cells got from the brick and tile unconscious processs. Table 1 Descriptive Statistics of brick and tile operation workers percentage alter cells shifting N N* Mean SE Mean St Dev. Minimum Q1 Median Q3 supreme % alter cells of 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 shamed cells of the workers in the brick industry is high(prenominal) when compared with the tile operation workers. The median percentage of brick industry workers is 1. 370 which i s higher as compared to the brick operation workers which is 1. snow. The inter-quartile range which creation the departure betwixt Q3 and Q1 is higher for the brick operation compared to that of the tile.Figure 1Box plot displaying %damage of cell in workers from both tile and brick industries. The figure above armys that the percentage-discredited cell for tile operators is overthrow when compared with the brick operators indicating a discrimination in the mean and median. Figure 1 shows a deflexion in the health disaster of the tile and brick workers. There is distinguish of lopsidedness in the crack uping of brick operators whereas the tile statistical statistical distri providedion is symmetric, as the median clientele for the brick operators has shifted outside(a) from the centre.The % cell damage in workers of the tile operation is closely grouped unconnected from the 2 extreme outliers when compared to the % cell damage of the brick workers, which is quite wid e. For the above box plot the urgency for a further analysis is to be carried out as the guess fag end non either be accepted neither deflected since the box plot tho denotes statistical measures (mean, median, Q1, Q3, max & min prizes) which are not ample to prove the difference amongst the two sites. Figure 2 Histogram of the Tile and Brick operation data The % of modify cells of the brick operation is higher when compared to the tile operation.This is think from the histogram above which exhibits that the bar revalues which is the % shamed cells for brick operation is higher than the bar value of the tile operation. We have used a histogram, as it is one of the important tools for a data analysis. Figure 3The experiment 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 trading trading trading operations-value obtained from the Levenes Test is 0. 200 which is also higher than 0. 05 implies that the guess of difference cannot be repudiateed.The value of the F-Test is 0. 952 which being higher than 0. 05 shows also shows no signs that the fruitless hypothesis (H0) should be spurned and also that thither is no difference between %cell damage of workers from brick and tile operations. The obtained values from the seek for equal variance point out to an abnormal distribution of data stating the acceptance of the null hypothesis. thence no clear consequence of a difference in the median among the % damaged cells in the workers of both the operations. Figure 4Normal Distribution Graph 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 haped a skewed distribution and it can also be seen that the a bove graph does not follow any trend or pattern. The is no win over 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 reason out that the residuals do not follow a normal distribution.A MANN WHITNEY TEST will be used to statistically analyse the data as the %damaged cells of workers in the tile operation shows that the data is not normally distributed since the P-Value is lower than 0. 05 and also that the plots on the graph so no way of life any precise trend. MANN WHITNEY TEST Results & CI Of Tile & Brick Manufacturing Operations Table 2illuminates the number of samples used in the Mann Whitney test and the obtained median for data of brick and tile manufacturing operations Sample type Number of sample Median Tile 27 1. ascorbic acidBrick 38 1. 370 Point 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. Contrariwise the difference in the median is 0. 200(estimated), which means that 0. 200%(approximately) more % of damaged cells in workers of the brick operations than those of the tile operations.A 100% veritable analysis cannot be proved as the confidence interval (CI) is just 95%, hence creating a need for more data in differentiate 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. Therefore it can be reason out that thither is no convincing evidence of difference in the median between %damaged cells of workers in the 2 operations. resultA use of unhomogeneous graphs and descriptive statisti cs were used and inferred to decide if at that place 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 decision may be drawn from the these analyses that at that place is scarce evidence to suggest that on that point 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 length of service of workers in two operations is paired. N Mean StDev SE Mean % Damaged cells 65 1. 451 1. 095 0. 136 length 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 reason 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 showing the correlation between the % of cells damaged with a reversal line and the length of service in years. The predicted value for retroflexion is 17. 4%, which states the 17. 4% of the variability in the data is represented by the atavism model. This cannot be used to get prox values as the predictive value itself is very low. Pearsons correlation need to be conducted since the above scatter plot shows a minor positive standoff between the % damaged cells and the length of the service, but the damage of the cells in the future cannot be predicted.Pearsons Correlation 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 Correlation 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. wherefore a outcome 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 there is not sufficient evidence acquirable to say that there is a correlation. closure 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 genuine difference between the health hazard of the worker at the tile and brick operation.The second test c oncluded 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 of 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.