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Data Analysis: The Sun Coast Remediation Data Set

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 Tool pack 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.

Formulating hypotheses for Correlation between microns and average employee sickness per year.

The Pearson correlation coefficient is r = -0.71598 when rounded to 4 decimal places. It indicates there is a strong negative correlation between the two variables.

The value of the coefficient of determination, r2 is 0.5126. This means that the regression model explains about 88.2772% of the data variation.

Significance		Interpretation
α-value	0.05	1.89E-17< 0.05 Reject H01 and accept HA1. Microns and average sickness of employee per year have a correlation that is significant, statistically speaking.
p-value	1.89E-17

The response variable is the safety training expenditure is the Y variable. On the other hand, the predictor variable; lost time hours are the X.

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 (Glen, 2010). 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 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 (Glen, 2010).

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 (Glen, 2010).

The multiple R is evaluated to be about 0.5837. Since this is a multiple linear regression analysis, this value of Multiple R communicates that the relationship between the predictors and the response variable is moderately strong. As aforementioned in the previous analysis, R square in this case is also the coefficient of determination. Its value is 0.3407. The interpretation is that the regression model accounts for only 34.07% of the data variation (Glen, 2010). The regression is a poor fit.

Now, the significance F-value from the ANOVA test is

Significance		Interpretation
Alpha value	0.05	1.2E-132 < 0.05 Reject H03 and accept HA3. At least one regression coefficient is different from 0 (The regression fit is significant).
F-value	1.2E-132

The X2 predictor (for Chord length) has a p-value of 0.58. Now 0.58>0.5 thus X2 is not a significant predictor of Y (frequency). The p-value of the rest of the X variables are smaller than α-value thus, X1, X3, X4 and X5 significantly predict the response variable Y (frequency) (Glen, 2010).

References

Glen, S. (2010). "Excel Regression Analysis Output Explained." StatisticsHowTo.com: https://www.statisticshowto.com/probability-and-statistics/excel-statistics/excel-regression-analysis-output-explained/