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Running head: DESCRIPTIVE STATISTICS 0
Descriptive statistics
Stephener Baisey
Columbia Southern University
Data Analysis
Descriptive Data and Assumptions: Correlation
Frequency Distribution Table
PM size | Frequency |
0-1 | |
2-4 | 24 |
5-7 | 37 |
8-10 | 34 |
Sick Days | Frequency |
0-2 | |
4-7 | 61 |
8-9 | 30 |
10-12 | 11 |
Histogram
Descriptive Statistics Table
microns |
|
| sick day |
|
Mean | 5.65728155 | Mean | 7.126214 | |
Standard Error | 0.25560014 | Standard Error | 0.186484 | |
Median | Median | |||
Mode | Mode | |||
Standard Deviation | 2.59405814 | Standard Deviation | 1.892605 | |
Sample Variance | 6.72913764 | Sample Variance | 3.581953 | |
Kurtosis | -0.8521619 | Kurtosis | 0.124923 | |
Skewness | -0.37325713 | Skewness | 0.14225 | |
Range | 9.8 | Range | 10 | |
Minimum | 0.2 | Minimum | ||
Maximum | 10 | Maximum | 12 | |
Sum | 582.7 | Sum | 734 | |
Count | 103 | Count | 103 | |
Largest(1) | 10 | Largest(1) | 12 | |
Smallest(1) | 0.2 | Smallest(1) | ||
Confidence Level (95.0%) | 0.50698167 |
| Confidence Level (95.0%) | 0.36989 |
Kolmogorov-Smirnov Test
The hypotheses used are:
Ho: The sample data provided has no significant difference to the data that relates to normal population.
H1: There is a significant difference that emerges between the sample data to that of normal population.
Use an alpha of .05 and provide the test statistic and p level here
P > 0.05
P ≤ 0.05
Accept or reject the null hypothesis here.
The null hypothesis is rejected
Measurement Scale
Ordinal
Measure of Central Tendency
Mean
Evaluation
The above descriptive statistics has indifferences as the test static of the sample data to that of normal population were different.
Assumptions for parametric testing
The assumptions in the parametric testing were not met as there was indifferences in the results under a 95 percent confidence interval. First there was differences in the data which led to differences in the measures of central tendency. For instance, the mean of the data for microns and sick day as projected by that of 5.65 and that of 7.12 respectively. Despite having similar counts that was also a difference that arose between the highest and lowest number in the data provided. Additionally, parameters in the test static for the two populations gave contrastive results. Thus, the assumptions in the parametric testing remained unmet.
Descriptive Data and Assumptions: Simple Regression
Frequency Distribution Table
Expenditure | Frequency |
20-500 | 108 |
501-1000 | 76 |
1001-1500 | 27 |
1501-2000 | 11 |
2001-2500 |
Time | Frequency |
0-50 | |
51-100 | 26 |
101-200 | 98 |
201-300 | 85 |
301-400 |
Histogram
Descriptive Statistics Table
safety training expenditure |
|
| lost time hours |
|
Mean | 595.9843812 | Mean | 188.0045 | |
Standard Error | 31.4770075 | Standard Error | 4.803089 | |
Median | 507.772 | Median | 190 | |
Mode | 234 | Mode | 190 | |
Standard Deviation | 470.0519613 | Standard Deviation | 71.72542 | |
Sample Variance | 220948.8463 | Sample Variance | 5144.536 | |
Kurtosis | 0.444080195 | Kurtosis | -0.50122 | |
Skewness | 0.951331922 | Skewness | -0.08198 | |
Range | 2251.404 | Range | 350 | |
Minimum | 20.456 | Minimum | 10 | |
Maximum | 2271.86 | Maximum | 360 | |
Sum | 132904.517 | Sum | 41925 | |
Count | 223 | Count | 223 | |
Largest(1) | 2271.86 | Largest(1) | 360 | |
Smallest(1) | 20.456 | Smallest(1) | 10 | |
Confidence Level (95.0%) | 62.03197147 |
| Confidence Level (95.0%) | 9.465484 |
Kolmogorov-Smirnov Test
State null and alternative hypotheses for normality here.
H0: The sample data that relates to training expenditure is different to that of lost time hours.
H1: There is a significant difference value between the data in training expenditure and that of lost time hours.
Use an alpha of .05 and provide the test statistic and p level here
P > 0.05
P ≤ 0.05
Accept or reject the null hypothesis here.
We accept the null hypothesis
Measurement Scale
Nominal
Measure of Central Tendency
Median
Evaluation
The p value for both training expenditures and the lost time hours is exceedingly high.
Assumptions for parametric testing
The assumptions for parametric testing in the study prove to be met as expressed by the test statistic. First, there is a huge difference that emerges in the data between the training expenditure and the lost time hours. Findings from the statistical test indicates that the p value in both the training expenditure and lost time hours is exceeding high. However, an analysis of the data indicates that there lost time hours has a smaller confidence interval as opposed to that of training expenditure. Thus, the statistical tests proves the assumptions as there is a great difference that emerges in the two data sets.
Descriptive Data and Assumptions: Multiple Regression
Frequency Distribution Table
Decibel | Frequency |
100-106 | |
107-111 | 51 |
112-116 | 126 |
117-121 | 249 |
122-131 | 786 |
132-141 | 287 |
Histogram
Descriptive Statistics Table
Decibel |
|
|
|
Mean | 124.8359 |
Standard Error | 0.177945 |
Median | 125.721 |
Mode | 127.315 |
Standard Deviation | 6.898657 |
Sample Variance | 47.59146 |
Kurtosis | -0.31419 |
Skewness | -0.41895 |
Range | 37.607 |
Minimum | 103.38 |
Maximum | 140.987 |
Sum | 187628.4 |
Count | 1503 |
Kolmogorov-Smirnov Test
State null and alternative hypotheses for normality here.
H0: There is no relationship between the X and Y variables.
H0= H1=0
H1 ≠ 0
Use an alpha of .05 and provide the test statistic and p level here
P=0
P>0
Accept or reject the null hypothesis here.
Reject
Measurement Scale
Internal
Measure of Central Tendency
Mean
Evaluation
There is no direct relation between the variables.
Assumptions for parametric testing
The assumptions for parametric testing were unmet as it is evident that is no relationship between the variables. In such circumstances there is a null hypothesis for each variable an indication that the variables do not fit in the multiple regression equation. Since the variable do not have any relations there remains a standard error in the data. Since the null hypothesis was untrue there is less probability of obtaining a test statistic based on the data provided. This is because there are two variables at the expense of three. This makes the parametric assumptions to remain unmet as there is no clear relationship.
Descriptive Data and Assumptions: Independent Samples t Test
Frequency Distribution Table
Training | Frequency |
49-60 | 12 |
61-70 | 20 |
71-80 | 21 |
81-90 | |
91-100 |
Training | Frequency |
74-80 | 14 |
81-85 | 21 |
86-90 | 19 |
91-95 | |
96-100 |
Histogram
Descriptive Statistics Table
Prior Training |
|
| Revised Training |
|
Mean | 69.79032 | Mean | 84.77419 | |
Standard Error | 1.402788 | Standard Error | 0.659479 | |
Median | 70 | Median | 85 | |
Mode | 80 | Mode | 85 | |
Standard Deviation | 11.04556 | Standard Deviation | 5.192742 | |
Sample Variance | 122.0045 | Sample Variance | 26.96457 | |
Kurtosis | -0.77668 | Kurtosis | -0.35254 | |
Skewness | -0.0868 | Skewness | 0.144085 | |
Range | 41 | Range | 22 | |
Minimum | 50 | Minimum | 75 | |
Maximum | 91 | Maximum | 97 | |
Sum | 4327 | Sum | 5256 | |
Count | 62 | Count | 62 | |
Largest(1) | 91 | Largest(1) | 97 | |
Smallest(1) | 50 | Smallest(1) | 75 | |
Confidence Level (95.0%) | 2.805048 |
| Confidence Level (95.0%) | 1.31871 |
Kolmogorov-Smirnov Test
State null and alternative hypotheses for normality here.
H0=0
H1>0
Use an alpha of .05 and provide the test statistic and p level here
P≠ 0
Accept or reject the null hypothesis here.
Accept
Place detailed test data in the appendix.
Measurement Scale
Internal
Measure of Central Tendency
Mean
Evaluation
There is an indirect relationship between the sample data and the normal population.
Assumptions for parametric testing.
The assumptions were met. Statically test indicate that the probability test is lower than the p value. For instance, in the first data the p value is 2.8 whereas the second data has a p value of 1.31. The p value is greater than 0. This indicates that there is a indirect relationship of the data as evidenced by the p value. The dependent variables were normally distributed. Additionally, there are two groups which are independent to each other such as the test scores for the revised training and that of prior training. Therefore, there is an indirect relationship of the data provided.
Descriptive Data and Assumptions: Dependent Samples t Test
Frequency Distribution Table
Exposure | Frequency |
5-15 | |
16-25 | |
26-35 | 12 |
36-45 | 16 |
46-56 |
Exposure | Frequency |
5-15 | |
16-25 | |
26-35 | 11 |
36-45 | 17 |
46-56 |
Histogram
Descriptive Statistics Table
Pre-Exposure μg/dL |
|
| Post-Exposure μg/dL |
|
Mean | 32.8571429 | Mean | 33.28571 | |
Standard Error | 1.75230655 | Standard Error | 1.781423 | |
Median | 35 | Median | 36 | |
Mode | 36 | Mode | 38 | |
Standard Deviation | 12.2661458 | Standard Deviation | 12.46996 | |
Sample Variance | 150.458333 | Sample Variance | 155.5 | |
Kurtosis | -0.57603713 | Kurtosis | -0.65421 | |
Skewness | -0.42510965 | Skewness | -0.48363 | |
Range | 50 | Range | 50 | |
Minimum | Minimum | |||
Maximum | 56 | Maximum | 56 | |
Sum | 1610 | Sum | 1631 | |
Count | 49 | Count | 49 | |
Largest(1) | 56 | Largest(1) | 56 | |
Smallest(1) | Smallest(1) | |||
Confidence Level (95.0%) | 3.52324845 |
| Confidence Level (95.0%) | 3.581792 |
Kolmogorov-Smirnov Test
State null and alternative hypotheses for normality here.
Ho: u1=0
H1:u1≠0
Use an alpha of .05 and provide the test statistic and p level here
α=0.05
t=m1-m2/Sd/n
33.28571-32.8571429=0.4285671/1.7523
t=0.24457
Accept or reject the null hypothesis here.
Accept
Measurement Scale
Interval
Measure of Central Tendency
Mean
Evaluation
The null hypothesis is accepted as the null hypothesis is greater than 0.
Assumptions for parametric testing
The assumptions for parametric testing were met. This is because the data consisted of dependent variable which were continuous on a ratio basis. Additionally, the observations of the data collected were independent of one another. This is irrespective of the fact that dependent variables were normally distributed. A comparison of the two means indicates that there is statistical difference between the mean. In the data present the difference between the mean is 0.4285671. An evaluation of the statistical test indicates that the t-test is greater than the calculated test. The differences between the observed t test and the calculated t-test leads to acceptance of the null hypothesis.
Descriptive Data and Assumptions: ANOVA
Frequency Distribution Table
Air | Frequency |
1-3 | |
4-6 | |
7-9 | |
10-12 | |
12-15 |
Soil | Frequency |
5-7 | |
8-10 | 13 |
10-13 |
Water | Frequency |
1-3 | |
4-6 | 10 |
7-9 | |
10-12 |
Training | Frequency |
1-3 | |
4-6 | 16 |
7-9 |
Histogram
Descriptive Statistics Table
A = Air |
|
| B = Soil |
|
Mean | 8.9 | Mean | 9.1 | |
Standard Error | 0.684028 | Standard Error | 0.390007 | |
Median | Median | |||
Mode | 11 | Mode | ||
Standard Deviation | 3.059068 | Standard Deviation | 1.744163 | |
Sample Variance | 9.357895 | Sample Variance | 3.042105 | |
Kurtosis | -0.6283 | Kurtosis | 0.11923 | |
Skewness | -0.36085 | Skewness | 0.492002 | |
Range | 11 | Range | ||
Minimum | Minimum | |||
Maximum | 14 | Maximum | 13 | |
Sum | 178 | Sum | 182 | |
Count | 20 | Count | 20 | |
Largest(1) | 14 | Largest(1) | 13 | |
Smallest(1) | Smallest(1) | |||
Confidence Level(95.0%) | 1.431688 |
| Confidence Level(95.0%) | 0.816294 |
C = Water |
|
| D = Training |
|
Mean | Mean | 5.4 | ||
Standard Error | 0.575829 | Standard Error | 0.265568 | |
Median | Median | |||
Mode | Mode | |||
Standard Deviation | 2.575185 | Standard Deviation | 1.187656 | |
Sample Variance | 6.631579 | Sample Variance | 1.410526 | |
Kurtosis | -0.23752 | Kurtosis | 0.253747 | |
Skewness | 0.760206 | Skewness | 0.159183 | |
Range | Range | |||
Minimum | Minimum | |||
Maximum | 12 | Maximum | ||
Sum | 140 | Sum | 108 | |
Count | 20 | Count | 20 | |
Largest(1) | 12 | Largest(1) | ||
Smallest(1) | Smallest(1) | |||
Confidence Level (95.0%) | 1.205224 |
| Confidence Level (95.0%) | 0.55584 |
Kolmogorov-Smirnov Test
State null and alternative hypotheses for normality here.
H0: There is no difference of the means.
H1: Means are not all equal
Use an alpha of .05 and provide the test statistic and p level here
α=0.05
Test statistic of water to training
(7-5.4)2=2.56
Test statistic of air to soil
(8.9-9.1)2=0.0004
Accept or reject the null hypothesis here.
Accept
Measurement Scale
Ratio
Measure of Central Tendency
Mean
Evaluation
The means are not equal as they base on different sets of data.
Assumptions for parametric testing
Based on the data provided the assumptions that can be derived are those for normality, equal variance and that of independent errors. From the data there is an interaction of the variables with no restrictions. The parametric assumptions in this scenario would relate to the parameters on the population distribution upon which data is drawn. Additionally, a non-parametric test would refer to that which makes no such assumptions. This leads to normal distribution, homogeneity of the variances, multiple groups which relates to the same variance as well as linearity on the independent relationships. Thus, the assumptions define the type of variance.
References
Judd, C. M., McClelland, G. H., & Ryan, C. S. (2017). Data analysis: A model comparison approach to regression, ANOVA, and beyond. Routledge.
Kalaian, S. A., & Kasim, R. M. (2016). Analyzing quantitative data. In Mixed Methods Research for Improved Scientific Study (pp. 149-164). IGI Global.