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Data Analysis: The Sun Coast Remediation Data Set

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Data Analysis: Hypothesis Testing

In this project, we are going to use the data set: Sun Coast Remediation in Microsoft Excel using the Data Analysis Toolpack to explore the correlation of variables and conduct regression analysis. The results of the analysis will be displayed here directly from Microsoft Excel and the resulting predictive regression equations will be discussed.

Correlation: Hypothesis Testing

Hypotheses:

i. Microns versus mean annual sick days per employee

Ho1: There is no significant linear relationship/correlation between microns and mean annual sick days per employee.

Ha1: There is a significant linear relationship/correlation between microns and mean annual sick days per employee.

	microns	mean annual sick days per employee
microns	1
mean annual sick days per employee	-0.715984185	1

Regression Statistics
Multiple R	0.715984185
R Square	0.512633354
Adjusted R Square	0.507807941
Standard Error	1.327783455
Observations	103

ANOVA
	df	SS	MS	F	Significance F
Regression	1	187.2953239	187.3	106.236	1.89059E-17
Residual	101	178.0638994	1.763
Total	102	365.3592233
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	10.08144483	0.315156969	31.989	1.17E-54	9.456258184	10.70663	9.456258	10.70663148
microns	-0.522376554	0.050681267	-10.307	1.89E-17	-0.62291455	-0.42184	-0.62291	-0.421838554

The Pearson correlation coefficient is r = -0.71598 when rounded to 4 decimal places.

Interpretation: It indicates there is a strong negative correlation between the two variables.

The value of the coefficient of determination, r2 is 0.5126.

Interpretation: About 51.26% of the variation between microns and mean annual sick days per employee is explained by the relationship.

From the results the p-value is 1.89E-17, a very small value. By using the alpha level of significance to be 0.05 then the p-value is less than the alpha value i.e., 1.89E-17 < 0.05. As a result, we reject the null hypothesis and accept the alternative hypothesis. Therefore, we conclude that there is a statistically significant linear relationship between mean annual sick days per employee

Simple Regression: Hypothesis Testing

Hypotheses:

Ho2: β1 = 0 (The regression is not significant)

Ha2: β1 ≠ 0 (The regression is significant)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.939559
R Square	0.882772
Adjusted R Square	0.882241
Standard Error	161.303
Observations	223
ANOVA
	df	SS	MS	F	Significance F
Regression	1	43300521	43300521	1664.211	7.7E-105
Residual	221	5750122	26018.65
Total	222	49050644

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	1753.602	30.36296	57.75465	2.6E-135	1693.764	1813.44	1693.764	1813.44
lost time hours	-6.15739	0.150936	-40.7947	7.7E-105	-6.45485	-5.85994	-6.45485	-5.85994

Y = safety training expenditure and X = lost time hours

The regression model is given by: Y = 1753.602 - 6.15739*X2. This means that for every additional hour in lost time hours, the safety training expenditure decreases by 6.15739 money units.

The multiple R is 0.939559. Since this is a simple linear regression analysis, the multiple R is same as correlation coefficient, r. As such, it indicates there is a strong positive correlation between safety training expenditure and lost time hours. R square is the coefficient of determination. Its value is 0.882772. Its interpretation is that the regression model explains 88.2772% of the data variation. The regression is a good fit.

The alpha level is 0.05. It is the level of significance. From the ANOVA results, the ANOVA F-value is 1664.210687. This is the ratio of mean sum of squares total (MST) to the mean sum of squares due to error (MSE). The significance F is 7.6586E-105, which is a very small value. Since 7.6586E-105 < 0.05, we reject the null hypothesis and accept the alternative hypothesis. We conclude that the regression fit is significant. The statistical significance of the X variable is also 7.6586E-105 which is less than 0.05. It means the lost time hours is a significant predictor of safety training expenditure.

Multiple Regression: Hypothesis Testing

Hypotheses:

Ho3: β1 = β2 = β3 = β4 = β5 = 0 (The regression fit is not significant)

Ha3: At least one is different from zero (The regression fit is significant)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.583706496
R Square	0.340713274
Adjusted R Square	0.338511248
Standard Error	2564.049485
Observations	1503

ANOVA
	df	SS	MS	F	Significance F
Regression	5	5.09E+09	1.02E+09	154.7271	1.2E-132
Residual	1497	9.84E+09	6574350
Total	1502	1.49E+10

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	32243.94	1307.24	24.67	5.27E-113	29679.72	34808.16	29679.72	34808.16
Angle in Degrees	-86.46	17.20	-5.03	5.581E-07	-120.20	-52.72	-120.20	-52.72
Chord Length	-741.56	1361.86	-0.54	0.5861673	-3412.92	1929.80	-3412.92	1929.80
Velocity (Meters per Second)	42.06	4.30	9.78	6.023E-22	33.63	50.50	33.63	50.50
Displacement	-65093.43	8026.09	-8.11	1.042E-15	-80837.01	-49349.86	-80837.01	-49349.86
Decibel	-241.11	10.27	-23.49	4.07E-104	-261.25	-220.97	-261.25	-220.97

Y = Frequency (Hz).

Ley X1 = Angle in Degrees, X1 = Angle in Degrees, X2 = Chord Length, X3 = Velocity, X4 = Displacement, X4 = Displacement, X5 = Decibel

The regression model is given by:

Y = 32243.94 - 86.46*X1 - 741.56*X2 + 42.06*X3 - 65093.43*X4 - 241.11*X6

For this model, Y is the response variable and the Xi’s, i = 1,2,3,4,5 are the predictor variables. If we pick the predictors one by one while each time holding all the others constant, then for every predictor variable with a negative coefficient we pick, the response variable (frequency) will decrease with the corresponding predictor coefficient units. The response variable will increase by the corresponding predictor variable coefficient units for every unit increase in the predictor variable if the coefficient of the predictor variable is positive while we hold all other predictor variables constant.

Multiple R value is 0.583706496. This indicates a moderately strong linear relationship between frequency and the predictor variables. The R^2 value is 0.340713274. This means that the model explains only 34.07% of the variability of the response data around its mean.

While the alpha level of significance is 0.05, the ANOVA significant F-value is 1.2E-132 which is very small compared to alpha = 0.05. We reject the null hypothesis and accept the alternative hypothesis. We conclude that the regression fit is significant. With the exception of the Chord Length predictor, the statistical significance of all the other predictor variables are very small when compared to the alpha = 0.05 level of significance. Chord length is therefore not a significant predictor of frequency. Angle, Velocity, Displacement and Decibel are all significant predictors of frequency.

References

Glen, S. (n.d). "Excel Regression Analysis Output Explained." StatisticsHowTo.com: https://www.statisticshowto.com/probability-and-statistics/excel-statistics/excel-regression-analysis-output-explained/