Discussion # III - It is important to remember that typically a two-factor regression model cannot accurately describe the entire situation.Look at the dependent variable in part 3. Name at least 2 independent factors you would use to run a Multiple Linear Regression (MLR)and explain why you feel they are related. Then use those factors to run a Multiple Linear Regression (MLR) on the data and see if thevariables you chose are related to the dependent variable in part 3 chose. What is your MLR equation? Is your MLR significant? Are any of the Independent factors significant? What is theR²value? Explain and interpret this value and how it relates to the MLR. Make sure you include your MLR Excel output as an attachment in yourresponse post.

Part 3, Table 3

Vehicle Type/Class	Year	Make	Model	Price	MPG (City)	MPG (Highway)	Cylinders
SUV	2022	Ford	Bronco, Sport	$28,760	25	28	3
SUV	2022	Jeep	Compass	$26,490	22	22-30	4
SUV	2022	Honda	HR-V	$25,845	28	34	4
SUV	2022	Nissan	Rogue, Sport	$25,485	25	32	4
SUV	2022	Chevrolet	Trailblazer	$24,795	26	30	3
SUV	2022	Toyota	RAV4	$27,700	25	32	4
SUV	2022	Audi	Q5	$45,195	23	28	4
SUV	2022	Mercedes-Benz	GLC	$44,900	18-22	24-28	4
SUV	2022	Toyota	4Runner	$43,765	16	19	6
SUV	2022	Kia	Seltos	$23,665	27	31	4

From the vehicle comparison data in table 3, I want to see if there is a correlation between MPG (Highway) and price of vehicle. In some instances, the price of a car is higher because it gets good gas mileage. I am assuming the better (or the higher) MPG, the higher the value of the car. I am expecting a positive correlation, but I want to see if my assumption is correct and what the actual correlation value is. My x-value, or independent variable, will be the MPG value since it is predicting the y-value, or the price of the car.

Price	MPG (Highway)
Observation 1 $28,760	28
Observation 2 $26,490	30
Observation 3 $25,845	34
Observation 4 $25,485	32
Observation 5 $24,795	30
Observation 6 $27,700	32
Observation 7 $45,195	28
Observation 8 $44,900	28
Observation 9 $43,765	19
Observation 10 $23,665	31

With the data, to find the correlation we will use the =CORREL( ) function in Excel. I got = -0.6945(rounded to the 4^thdecimal).

Next, to find r², or the coefficient of determination, I am going to square the correlation which gives me 48.23%. A 48.23% of variation in the data between MPG and price of car may not give us a good indication of what the data will look like, but I went on and interpreted the model further.

Regression using Excel.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.694493
R Square	0.48232
Adjusted R Square	0.417611
Standard Error	6914.492
Observations	10
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3.56E+08	3.56E+08	7.453575	0.025841
Residual	8	3.82E+08	47810195
Total	9	7.39E+08
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	76428.84	16543.24	4.619945	0.00171	38280.07	114577.6	38280.07	114577.6
MPG (Highway)	-1533.18	561.5787	-2.73012	0.025841	-2828.18	-238.177	-2828.18	-238.177
RESIDUAL OUTPUT
Observation	Predicted Price	Residuals	Standard Residuals
1	33499.82	-4739.82	-0.72707
2	30433.46	-3943.46	-0.60491
3	24300.74	1544.261	0.236885
4	27367.1	-1882.1	-0.28871
5	30433.46	-5638.46	-0.86492
6	27367.1	332.9024	0.051066
7	33499.82	11695.18	1.794003
8	33499.82	11400.18	1.748751
9	47298.43	-3533.43	-0.54202
10	28900.28	-5235.28	-0.80307

The p-value associated with this model is0.025841. Since the p-value

Next,using the equation =1+0, the regression equation and replace "x" and "y" with the actual variable names.of car=-1533.18(MPG Highway) +76428.84

This means when MPG Highway equals 0, the Price of a Car should be $76428.84.However,this does make sense as it does not have a practical meaning does not always have a practical meaning. The MPG Highway for a car will never be 0. So, the y-intercept is not meaningful and would not have a practical meaning in the problem.