Question

Provide justification for why you selected those variables. Run regression and explain your results and summarize your findings.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.311223884
R Square	0.096860306
Adjusted R Square	0.037959891
Standard Error	154.0999081
Observations	50
ANOVA
	df	SS	MS	F	Significance F
Regression	3	117153.0224	39051.00748	1.644475786	0.192145339
Residual	46	1092351.958	23746.78169
Total	49	1209504.98
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	189.8626807	48.2878658	3.931892155	0.000281818	92.66424818	287.0611133	92.66424818	287.0611133
number_of_reviews	-0.365674155	0.206843839	-1.767875501	0.083713991	-0.78202921	0.050680898	-0.782029208	0.050680898
minimum_nights	-0.575243694	0.872441711	-0.659349143	0.512959124	-2.33137778	1.180890386	-2.331377775	1.180890386
availability_365	0.170777604	0.167614247	1.01887284	0.313592312	-0.16661238	0.508167585	-0.166612377	0.508167585

Answer 1

Based on the regression output summary,

Let the significance level = 0.05

The results can be interpreted in following points,

1)

Overall Significance

	F	Significance F
Regression	1.644476	0.192145339

The significance F value is 0.192145339 which is greater than 0.05 at 5% significance level which mean the model doesn't fit the data value at the predefined significance level. Hence we can conclude that independent variables doesn't fit the model significantly.

2)

Significance of Independent variables

From, the result summary,

	Coefficients	t Stat	P-value
number_of_reviews	-0.365674155	-1.767875501	0.083714	>	0.05	Not Significant
minimum_nights	-0.575243694	-0.659349143	0.512959	>	0.05	Not Significant
availability_365	0.170777604	1.01887284	0.313592	>	0.05	Not Significant

The P-value for each independent variable is greater than 0.05 at 5% significance level hence we can conclude that independent variables are not significant in the model.

3)

R-Square value

From, the result summary,

R Square

0.096860306

The R-square value tell, how well the regression model fit the data values. The R-square value of the model is 0.096860306 which means, the model explains approximately 9.686% of the variance of the data value. Based on this evidence we can conclude the model is not a good fit.

Conclusion As the linear regression model is not a good fit, you can try to fit some nonlinear function of predictor variable by examining the display of the predictor variable vs response variable.