Minitab Project Essay Sample

Introduction

In this paper we will show basics of use of Mintab 16 statistical software related to a real world problem. As it mentioned in the case study, “fine dusts formed during production activities are a potential health hazard for those who are working in brick, tile, porcelain and crystal glass manufacturing. One of the reasons for this is that the smallest particles in fine dusts may contain Respirable crystalline silica formed from the quartz component of the raw materials.”
In this case we are given with a data set of workers in brick, tile, porcelain and crystal glass manufacturing . We will examine how the percent of damaged cells is related to the age of the worker, his length of service and sector of manufacturing. Starting with a descriptive and graphical statistics for each variable, the appropriate test then performed to check all hypotheses. The results of the tests are interpreted and the conclusions and recommendations given.

Descriptive Statistics

The set of workers contains of 127 observations of 5 variables. According the conditions, the list of the variables and their meaning is shown below:

“C1 - A worker ID, identifying the sector (letter B,T, P or C) and the individual (3 digit number)

C2 - The sector in which the employee worked (brick, tile, porcelain or crystal glass)
C3 - Length of service of each employee in the sector (in years)
C4 - Age of each employee (in years)
C5 – The percentage of damaged cells for each employee as calculated by results of the LDH assay (%)”
In this assignment we would like to investigate which factors are lead to the growth of percentage of damaged cells. Is there a difference in this percentage between the sector groups? Does the age has a significant effect on percentage of damaged cells or the length of service is a main factor? All these questions will be discussed.

The table below shows the Minitab 16 output about 3 quantitative variables (C3, C4 and C5):

Descriptive Statistics: length of service (years); age (years); % damaged cells
Variable N N* Mean SE Mean StDev Variance Minimum
length of service (years 127 0 8,798 0,565 6,368 40,556 0,540
age (years) 127 0 39,00 1,02 11,44 130,91 17,00
% damaged cells 127 0 1,7098 0,0934 1,0525 1,1077 0,2000

Variable Q1 Median Q3 Maximum Range IQR

length of service (years 3,707 8,000 11,347 34,000 33,460 7,640
age (years) 30,00 38,51 47,00 65,00 48,00 17,00
% damaged cells 0,9647 1,6059 2,3000 4,7000 4,5000 1,3353

N for

Variable Mode Mode
length of service (years 1 8
age (years) 40; 47 4
% damaged cells 1,1 6
The average value of length of service is 8.798. The standard deviation of this variable is 6.368 which is quite large indicator. The data of length of service is very dispersed around the mean.
The most typical value of age is 39, with a standard deviation of 11.44. The data is not much dispersed around the mean.
These descriptives may be visualized on a histogram. The visual representation helps to better understand the distribution of the variables.
A blue bell-shaped curve on each graph is a normal distribution curve. It is added for comparison of distribution of each variable with a normal distribution. It seems that all three variables have the distribution quite close to normal. That’s why the data set may be considered as good.
However, the general descriptive statistics is not very informative. The one of the goals for this assignment is to investigate the difference in percentage of damaged cells depending on the sector of working. That’s why we produce descriptive statistics by the factor of sector:
Descriptive Statistics: length of service (years); age (years); % damaged cells

Variable sector Mean SE Mean StDev Minimum Median

length of service (years brick 8,51 1,22 7,52 0,54 6,44
crystal glass 7,113 0,741 4,056 0,888 7,017
porcelain 9,853 0,977 5,528 2,467 9,692
tile 9,83 1,43 7,45 1,00 8,29
age (years) brick 38,05 2,19 13,53 18,00 34,50
crystal glass 38,21 1,55 8,50 19,84 37,96
porcelain 38,60 1,74 9,87 23,13 38,76
tile 41,70 2,49 12,94 17,00 42,00
% damaged cells brick 1,572 0,185 1,142 0,200 1,370
crystal glass 2,108 0,186 1,017 0,300 2,000
porcelain 1,814 0,150 0,847 0,489 1,855
tile 1,337 0,205 1,065 0,200 1,100

Variable sector Maximum

length of service (years brick 34,00
crystal glass 16,398
porcelain 25,000
tile 27,23
age (years) brick 65,00
crystal glass 61,00
porcelain 64,00
tile 65,00
% damaged cells brick 4,300
crystal glass 4,700
porcelain 3,676
tile 4,500
The most long time working is on average in porcelain and tile sector and the less time is in crystal glass sector. The older workers are in tile sector, the average age of workers in other three sectors is almost equal. It seems that the highest percent of damaged cells is in crystal glass manufacturing, then goes porcelain sector, brick sector, then tile. The measure of variation for each sector is close to each other (the standard deviation is around 1 for all 4 sectors).

Hypothesis Testing

The first hypothesis we want to test is: if there is a significant difference in percentage of damaged cells between the sectors of working?
H0:μ1=μ2=μ3=μ4Ha:not all means are equal

Set level of significance alpha at 0.05:

α=0.05

One-way ANOVA: % damaged cells versus sector

Source DF SS MS F P
sector 3 9,59 3,20 3,02 0,032

Error 123 129,98 1,06

Total 126 139,57
S = 1,028 R-Sq = 6,87% R-Sq(adj) = 4,60%

Individual 95% CIs For Mean Based on

Pooled StDev
Level N Mean StDev ------+---------+---------+---------+---
brick 38 1,572 1,142 (-------*--------)
crystal glass 30 2,108 1,017 (---------*--------)
porcelain 32 1,814 0,847 (--------*--------)
tile 27 1,337 1,065 (--------*---------)
------+---------+---------+---------+---

1,20 1,60 2,00 2,40

Pooled StDev = 1,028
The next hypothesis to check is that if there is an association between each pair of the following variables: lengths of working, age, percentage of damaged cells. Start with correlation analysis:
H0: ρ=0Ha: ρ≠0a=0.05

Correlations: length of service (years); age (years); % damaged cells

length of service age (years)
age (years) 0,538

0,000

% damaged cells 0,460 0,284

0,000 0,001

Cell Contents: Pearson correlation
P-Value
The last step of this analysis is to develop a multiple linear regression equation for the response variable (% of damaged cells) predicted by age and length of service.
H0:β0=β1=β2=0Ha:not all beta are 0a=0.05

Regression Analysis: % damaged ce versus length of se; age (years)

The regression equation is
% damaged cells = 0,899 + 0,0716 length of service (years) + 0,00465 age (years)

Predictor Coef SE Coef T P

Constant 0,8989 0,3011 2,99 0,003
length of service (years) 0,07157 0,01562 4,58 0,000
age (years) 0,004646 0,008693 0,53 0,594
S = 0,940804 R-Sq = 21,4% R-Sq(adj) = 20,1%

Analysis of Variance

Source DF SS MS F P
Regression 2 29,820 14,910 16,85 0,000
Residual Error 124 109,754 0,885
Total 126 139,574
Results and Interpretation
According to One-Way ANOVA output we conclude the following:
Since p-value of the test is 0.032 and it is lesser than level of significance alpha, we reject the null hypothesis. There is enough evidence to say that the percentage of damaged cells is significantly different between the sectors of manufacturing (at 5% level of significance).
The correlation analysis shows that all correlations between the variable are significant. However, they are not strong. There is a positive moderate linear association between length of service and age (and this is natural and actually obvious). The association between percentage of damaged cells and length of service is also moderate positive linear and the association between age and percentage of damaged cells is positive linear and weak.
Since age and length of working have some impact on the percentage of the damaged cells (see correlation analysis result mentioned above), it has some sense to try to construct a linear regression. However, the obtained regression equation is not good. Even if according the ANOVA it is significant (F=16.85, p<0.001) and the coefficients are also significant, the R-squared is reported at only 21.4%. This means that this equation explains only 21.4% of the variance of the response variable. This indicator is too low.

Conclusions and Recommendations

The most useful result gives One-Way ANOVA. We have found that there is a difference in percentage of damaged cells between the sectors of manufacturing. The individual 95% confidence intervals were constructed for each variable and it seems that the most harmful sector is crystal glass manufacturing. This is might be the explanation why the average length of service in crystal glass manufacturing is the shortest among all four sectors. The least harmful is tile manufacturing, where workers serve for 9.83 years on average (almost the longest time among four sectors). It is recommended for the manager of the company to pay attention to this difference and, hence, make some compensation programs for workers depending on the sector of manufacturing.