Using the Sun Coast data set, perform a correlation analysis, simple regression analysis, and multiple regression analysis, and interpret the results.

Please follow the  

Unit V Scholarly Activity template

(ATTACHED) to complete your assignment.

You will utilize Microsoft Excel ToolPak for this assignment.

Example:

I. Correlation Analysis

a. Restate the hypotheses.

b. Provide data output results from Excel Toolpak.

c. Interpret the correlation analysis results   

II. Simple Regression Analysis

d. Restate the hypotheses.

e. Provide data output results from Excel Toolpak.

f. Interpret the simple regression analysis results       

III. Multiple Regression Analysis

g. Restate the hypotheses.

h. Provide data output results from Excel Toolpak.

i. Interpret the multiple regression analysis results.       

The title and reference pages do not count toward the page requirement for this assignment. This assignment should be no less than two pages in length, follow APA-style formatting and guidelines, and use references and citations as necessary.

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InsertTitle Here

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Course Name Here

Instructor Name Here

Date

Data Analysis: Hypothesis Testing

Use the Sun Coast Remediation data set to conduct a correlation analysis, simple regression analysis, and multiple regression analysis using the correlation tab, simple regression tab, and multiple regression tab respectively. The statistical output tables should be cut and pasted from Excel directly into the final project document. For the regression hypotheses, display and discuss the predictive regression equations.

Correlation: Hypothesis Testing

Restate the hypotheses:

Example:

Ho1: There is no statistically significant relationship between height and weight.

Ha1: There is a statistically significant relationship between height and weight.

Enter data output results from Excel Toolpak here.

Interpret and explain the correlation analysis results below the Excel output. Your explanation should include: r, r2, alpha level, p value, and rejection or acceptance of the null hypothesis and alternative hypothesis.

Example:

The Pearson correlation coefficient of r = .600 indicates a moderately strong positive correlation. This equates to an r2 of .36, explaining 36% of the variance between the variables.

Using an alpha of .05, the results indicate a p value of .023 < .05. Therefore, the null hypothesis is rejected, and the alternative hypothesis is accepted that there is a statistically significant relationship between height and weight.

Note: Excel data analysis Toolpak does not automatically calculate the p value when using the correlation function. As a workaround, the data should also be run using the regression function. The Multiple R is identical to the Pearson r in simple regression, R Square is shown, and the p value is generated. Be sure to show your results using both the correlation function and simple regression function.

Simple Regression: Hypothesis Testing

Restate the hypotheses:

Ho2:

Ha2:

Enter data output results from Excel Toolpak here.

Interpret and explain the simple regression analysis results below the Excel output. Your explanation should include: multiple R, R square, alpha level, ANOVA F value, accept or reject the null and alternative hypotheses for the model, statistical significance of the x variable coefficient, and the regression model as an equation with explanation.

Multiple Regression: Hypothesis Testing

Restate the hypotheses:

Ho3:

Ha3:

Enter data output results from Excel Toolpak here.

Interpret and explain the simple regression analysis results below the Excel output. Your explanation should include: multiple R, R square, alpha level, ANOVA F value, accept or reject the null and alternative hypotheses for the model, statistical significance of the x variable coefficients, and the regression model as an equation with explanation.

References

Include references here using hanging indentations. Remember to remove this example.

Creswell, J. W., & Creswell, J. D. (2018). Research design: Qualitative, quantitative, and mixed methods approaches (5th ed.). Sage.

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DataAnalysis: The Sun Coast Remediation Data Set

Insert Your Name Here

Insert University Here

Course Name Here

Instructor Name Here

Date

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

1

microns

mean annual sick days per employee

1

– 0.715984185

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

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microns

		ANOVA
							df			SS			MS			F			Significance F
		Regression	187.2953239	187.3	106.236	1.89059E-17
		Residual	101	178.0638994	1.763
		Total	102	365.359 223 3
		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	9.456258 184	10.70663	9.456258	10.70663148
-0.522376554	0.050681267	-10.307	1.89E-17	-0.62291 455	-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)

Regression Statistics

Multiple R

R Square

Adjusted R Square

Standard Error

Observations

ANOVA

 

df

SS

MS

F

Significance F

Regression

1

43300521

Residual

Total

 

 

 

	SUMMARY OUTPUT
0.939559
0.882772
0.882241
161.303
223
	43300521	1664.211		7.7E-105
221	5750122	26018.65
222	49050644

 

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

1693.764

1813.44

7.7E-105

-6.45485

-5.85994

1753.602	30.36296	57.75465	2.6E-135		1693.764		1813.44
lost time hours	-6.15739	0.150936	-40.7947		-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

R Square

Adjusted R Square

Standard Error

Observations

0.583706496

0.340713274

0.338511248

2564.049485

1503

ANOVA

 

df

SS

MS

F

Significance F

Regression

Residual

Total

 

 

 

5	5.09E+09	1.02E+09	154.7271	1.2E-132
1497	9.84E+09	6574350
1502	1.49E+10

 

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

29679.72

34808.16

-120.20

-52.72

-3412.92

1929.80

33.63

50.50

-80837.01

-49349.86

-261.25

-220.97

32243.94	1307.24	24.67	5.27E-113		29679.72		34808.16
Angle in Degrees	-86.46	17.20	-5.03	5.581E-07		-120.20		-52.72
Chord Length	-741.56	1361.86	-0.54	0.5861673		-3412.92		1929.80
Velocity (Meters per Second)	42.06	4.30	9.78	6.023E-22		33.63		50.50
Displacement	-65093.43	8026.09	-8.11	1.042E-15		-80837.01		-49349.86
Decibel	-241.11	10.27	-23.49	4.07E-104		-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/