All posts must 100% original work. no plagiarism. Post results must be provided using Excel. MAKE SURE YOU INTERPRET YOUR RESULTS ON A WORD DOCUMENT LISTING EACH CALCULATION BY STEPS FOR PEERS TO UNDERSTAND THE CALCULATION. Attach is the Forum 2 excel for the calculation.
Look back to the raw data you collected in Forum 1 attached. There are 7 variables listed:
Vehicle type/class Year Make Model Price MPG (city) MPG (highway)
Choose TWO variables that you feel are correlated and explain why you feel that they are correlated. Do you suspect the relation is positive or negative? Why? Which would be considered the independent variable, which the dependent variable? Why?
Run a regression analysis in Excel and provide the results in your post along with your raw data. Looking at the R2 value, explain what this indicates about the strength of the relation. Then write out your Regression Equation, state if your p-value and conclusion.
I encourage you to review the Week 7 Regression PDF at the bottom of the discussions. This will give you a step by step example on how to calculate a correlation and run a Regression using Excel. I DO NOT recommend doing this by hand. Let Excel do the heavy lifting for you.
Recall, our Car Price data
Car Price: Year Years
Old
Observation 1 $ 20,000 2015 4
Observation 2 $ 25,000 2016 3
Observation 3 $ 30,000 2018 1
Observation 4 $ 31,000 2018 1
Observation 5 $ 22,500 2016 3
Observation 6 $ 25,000 2016 3
Observation 7 $ 29,500 2018 1
Observation 8 $ 24,000 2015 4
Observation 9 $ 24,500 2017 2
Observation 10 $ 25,000 2017 2
With the Regression output,
Next, we want to test the hypothesis and see if the results are significant.
The hypothesis scenario looks like:
Ho: 𝜌 = 0
Ha: 𝜌 ≠ 0
If we look at the p-value or the Significance F we see the p-value = .000673.
.000673 < .05, Yes this is significant. This means Years Old is a significant
predictor of the Price of a Car.
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.884606501
R Square 0.782528661
Adjusted R Square 0.755344744
Standard Error 1725.490814
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 85706451.61 85706451.61 28.78646 0.000673381
Residual 8 23818548.39 2977318.548
Total 9 109525000
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 31959.67742 1296.435244 24.65196589 7.83E-09 28970.09239 34949.26245 28970.09239 34949.26245
Years Old -2629.032258 490.0064638 -5.365301179 0.000673 -3758.98919 -1499.075326 -3758.98919 -1499.075326
We can also compare r, the correlation to a critical value. If r < negative critical
value or r > positive critical value, then r is significant. If r is significant and the
line may be used for prediction.
We know r = -.8846. There is a correlation critical value table in RealizeIT under
Testing the Hypothesis in Week 7 but here is a link to a more detailed table.
Correlation CV Table
We know alpha = .05, this is two tailed test from the hypothesis scenario and n =
10. The critical value that corresponds to this in the table is r CV = 0.632. We
know our correlation is negative, so we will use the negative value of this.
-.8846 < -0.632, this tells us that r is significant and you can use the line for
prediction.
This is the same conclusion we got with the p-value from above.
Lastly, we can run a t-test to see if the data is significant. From the regression
output the t-Stat for the slope is -5.3653. But if we didn’t have the regression
output we can calculate this value using this equation.
t =
𝑟√𝑛−2
√1−𝑟2
Plugging in our correlation and sample size we get:
t =
−.8846√10−2
√1−(−.8846)2
=
− 2.50202
.4663505
= −5.3651
t – Test Stat we calculated by hand is very close to the t-stat in the output. It is a
little off because I did round some of my values.
Then we can use the =T.DIST.2T function to find the p-value. This Excel function
should look familiar.
=T.DIST.2T(ABS(-5.3651),8)
Remember if you have a negative value you will need to use the ABS function to
take the absolute value of it.
p-value = 0.000673544 < .05, Yes, this is significant. This is the same conclusion as
we got above, and this is the same p-value from the Regression Output.
It does not matter what way you use to Test the Hypothesis of a Simple Linear
Regression example, if done correctly you will get the same conclusion every
time.
../../../ITT%20Tech/Fall%20’14/correlation%20CV%20table
Recall, our Car Price data
Car Price: Year Years
Old
Observation 1 $ 20,000 2015 4
Observation 2 $ 25,000 2016 3
Observation 3 $ 30,000 2018 1
Observation 4 $ 31,000 2018 1
Observation 5 $ 22,500 2016 3
Observation 6 $ 25,000 2016 3
Observation 7 $ 29,500 2018 1
Observation 8 $ 24,000 2015 4
Observation 9 $ 24,500 2017 2
Observation 10 $ 25,000 2017 2
With the Regression output,
Lastly, I want to use my Regression Equation to predict prices. And then we want
to find a 95% prediction interval for that predicted price.
What would I expect to pay for a car that was manufactured in 2014? Remember
2019 – 2014 = 5. This means the car is 5 Years Old. This is the value you want to
substitute into the Regression Equation. DO NOT put 2019 into the equation.
𝑃𝑟𝑖𝑐�̂� = −2,629.03 (𝑌𝑒𝑎𝑟𝑠 𝑂𝑙𝑑) + 31,959.68
𝑃𝑟𝑖𝑐�̂� = −2,629.03 (5) + 31,959.68
𝑃𝑟𝑖𝑐�̂� = −13,145.16 + 31,959.68
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.884606501
R Square 0.782528661
Adjusted R Square 0.755344744
Standard Error 1725.490814
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 85706451.61 85706451.61 28.78646 0.000673381
Residual 8 23818548.39 2977318.548
Total 9 109525000
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 31959.67742 1296.435244 24.65196589 7.83E-09 28970.09239 34949.26245 28970.09239 34949.26245
Years Old -2629.032258 490.0064638 -5.365301179 0.000673 -3758.98919 -1499.075326 -3758.98919 -1499.075326
𝑃𝑟𝑖𝑐�̂� = $18,814.52
In the Year 2014, when the car is 5 Years Old, we will expect to pay $18,814.52 for
a car.
Now that we know what are expected to pay for a 5-Year-Old car, lets calculate a
95% prediction interval for 5 Years.
We need to use this equation:
�̂� ± 𝑡∗(𝑆𝐸)√1 +
1
𝑛
+
(𝑥0 − �̅�)
2
(𝑛 − 1)𝑆𝐷𝑥
2
We will use the =T.INV.2T function to find the T-Critical Value. This value should
look familiar. DF = n – 2 = 10 – 2 = 8. Which is the same DF for the Residual in the
Regression output.
=T.INV.2T(0.05,8)
2.306004135
Next, we will need to calculate the mean and SD for the x-variable. You should
recall how to calculate descriptive statistics from Week 2.
Mean = 2.4
SD = 1.1737878
SE is the Standard Error from the Regression Output which is 1725.4908
Now we can plug in what we know
18814.52 ± 2.306(1725.4908)√1 +
1
10
+
(5 − 2.4)2
(10 − 1)1.17378782
18814.52 ± 3978.9817√1 + .1 +
6.76
12.4
18814.52 ± 3978.9817√1.64516129
18814.52 ± 3978.9817(1.28263841)
18814.52 ± 5103.59476
($13,710.93, $23,918.11)
The 95% prediction interval for a 5-Year-Old car will go from $13,710.93 to
$23,918.11.
MLR
Recall: Linear Regression is a data analysis technique that tries to find a linear
pattern in the data. We use all the data to calculate a straight line which may be
used to predict the values.
The equation of line for a Simple Linear Regression (SLR) is:
�̂� = 𝛽1𝑥 + 𝛽0
Where 𝛽1 is the slope coefficient or the coefficient, 𝛽0 is the y-intercept and �̂� is
the predicted y value.
For Multiple Linear Regression (MLR) the equation of a line will look like:
�̂� = 𝛽1𝑥1 + 𝛽2𝑥2 + 𝛽3𝑥3 + 𝛽4𝑥4+ 𝛽5𝑥5 + … + 𝛽0
Where 𝛽1, 𝛽2, 𝛽3, 𝛽4, 𝛽5, … are the slopes coefficient or the coefficients, 𝛽0 is the
y-intercept and �̂� is the predicted y value.
For Multiple Linear Regression you will have more than 1 slope, but you will still
only have 1 y-intercept. The number of slopes will depend on the number of x-
variables you have. In the example equation above, I only wrote out 5 slopes but
know you can have any amount as along as it is 2 and above. Because if there is
only 1-slope then this is a SLR NOT a MLR.
Example: Let’s move away some our car price data and look at home prices. Is
the MLR model a good predictor of home prices? If so, what variables help
predict the price of a home?
Data:
Price Area (Sq Ft) Floor Bedrooms Bathrooms
$ 69,000 600 1 2 1
$ 118,500 1000 2 2 2
$ 125,000 1100 1 3 2
$ 139,300 1300 2 3 2
$ 147,900 1700 2 3 2
$ 169,900 1800 1 3 2.5
$ 134,900 1300 1 4 2.5
$ 169,900 1700 2 4 3
$ 194,500 2000 2 5 3.5
$ 209,900 2100 3 5 4
Looking at our example, we want to use Area, Floors, Bedrooms and Bathrooms
to try and predict home prices. This means the x-variables are Area, Floors,
Bedrooms, and Bathrooms and the y-variable is Price.
Because there are 4 x-variables, this means there are 4 slopes in the MLR model.
Next, we will run a Regression using Excel. We will use the Data Analysis ToolPak
to run the Regression.
Go to Data – > Data Analysis
When the new window pops ups, scroll to where it says “Regression”, highlight it
and Click “OK”
Then it will say “Input”
Input Y Range: Click in the box and highlight the y values or the price column.
Input x Range: Click in the box and highlight the x values or the area, floors,
bedrooms, and bathrooms columns.
Check the box that say “Labels” this will tell you that the first row has labels in it.
Output Options
Make sure the second bubble is highlighted. “New Worksheet Ply”
Make sure you check the box for Residuals and Standardized Results
Then Click “OK”
(Remember, the x-values predicts the y-value. Area, Floors, Bedrooms and
Bathrooms will predict what the Price of a Home. This is very important to
understand and remember)
It should look like this:
Once you click OK, here is the Regression Output:
Looking at the output we see the estimated regression line is:
𝑃𝑟𝑖𝑐�̂� = .053181(𝐴𝑟𝑒𝑎) − .111766(𝐹𝑙𝑜𝑜𝑟) − 5.382699(𝐵𝑒𝑑) + 24.79927(𝐵𝑎𝑡ℎ) +
27.9676
To interpret the Bathroom Slope we state:
For holding all other slopes constant, if the number of Bathrooms in a house increases
by 1, then the price of the home will increase by $24,799.
This makes sense because bathrooms are something everyone uses, and a good selling
point for the home that we will discuss below some more.
The y-intercept is $27,968. This means if you have 0 Sq. Ft., 0 Floors, 0 Bedrooms and 0
Bathrooms then the price of a home will be $27,968. This doesn’t make sense in the
context of our problem because if you don’t have anything in a home, then you will have
a plot of land. And plots of land sell differently than homes that are already done, and
you would use different x-variables as predictors for plots of land.
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.994699234
R Square 0.989426567
Adjusted R Square 0.98096782
Standard Error 5.602053083
Observations 10
ANOVA
df SS MS F Significance F
Regression 4 14683.58101 3670.895252 116.9708249 3.99315E-05
Residual 5 156.9149937 31.38299874
Total 9 14840.496
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 27.96762856 7.053708269 3.964953964 0.010689849 9.835494214 46.09976291
Area (Sq Ft) 0.053181736 0.008231449 6.460798716 0.001322498 0.032022122 0.074341349
Floor -0.11176635 3.801943471 -0.029397162 0.977685138 -9.884973176 9.661440476
Bedrooms -5.382699832 4.706373686 -1.143704302 0.304530968 -17.48081854 6.715418878
Bathrooms 24.79927334 7.643770423 3.24437705 0.022837965 5.150335932 44.44821074
Is this model a good predictor for Price of a Home?
Looking at the overall Significance F value, we get 3.99315E-05.
The “E” means scientific notation in Excel. This can be rewritten as
3.99315 x 10-5, moving the decimal place 5 places to the LEFT the overall p-value is:
.0000399315. .0000399215 < .05, Yes, this model is a good predictor for Price of a
Home.
Now that we know the overall model is significant, which variables, or slopes, are a good
predictor for Home Price? We want to look at the individual p-values for each slope.
- Area: .001322 < .05, Yes this is significant and a good predictor for Home Price.
– Floor: .97769 > .05, No, this is not significant and not a good predictor for Home Price.
– Beds: .30453 > .05, No, this is not significant and not a good predictor for Home Price.
– Baths: .02284 < .05, Yes this is significant and a good predictor for Home Price.
We see that Area and Bathrooms are significant predictors for Home Price because
those p-values are less than .05.
From the Regression output we notice that the Standard Error is 5.602 and the Adjusted
R-squared is 98.1%.
The last thing we want to look at are the standardize residuals to see if there are any
outliers.
RESIDUAL OUTPUT
Observation Predicted Price Residuals Standard Residuals Price
1 73.79877725 -4.79877725 -1.149263527 69,000$
2 119.7589785 -1.25897848 -0.301513901 118,500$
3 119.8062186 5.193781442 1.24386344 125,000$
4 130.3307993 8.969200671 2.148042024 139,300$
5 151.6034936 -3.703493572 -0.886953043 147,900$
6 169.4330702 0.466929849 0.111825454 169,900$
7 137.4595025 -2.559502515 -0.612977585 134,900$
8 171.0200671 -1.120067077 -0.268245883 169,900$
9 193.9915246 0.508475405 0.121775237 194,500$
10 211.5975685 -1.697568474 -0.406552217 209,900$
Any Standardize Residuals outside of the range -2 to 2 can be considered an outlier.
Anything outside of -3 to 3, is also an outlier and you might want to do additional
research to investigate this data point.
When the Home Price is $139,300 the Predicted Price was $130,331. The Standardize
Residual is 2.14. This data point can be considered an outlier. This means that the
model underpredicted this data point for the house.
If you are buying this home, then this is good news for you, because you can make a
lower offer.
If you are selling this home, that might not be such good news for you because you want
to get as much money for your home as possible.
When the Home Price is $209,900 the Predicted Price was $211,596. The Standardize
Residual is -.406. This data point is not an outlier. But this means that the model
overpredicted this data point for the house.
If you are selling this home, this might be good news for you because then you can list
the home for higher than you originally thought. But you don’t want to list it too high
because you do want offers. But it is something to consider.
Make sure you look at the context of the problem to see if the outliers play into your
favor or not. But regardless, you will want to look at a range from -2 to 2.
The correlation is the direction and strength of association between 2 variables is
often expressed in a single number called the correlation coefficient. This is
denoted by the variable r.
• r can only be between -1 and 1, -1 ≤ r ≤ 1.
• If r = 0, then there is no linear relationship at all.
• If r = -1, then there is a perfect linear relationship that slopes down.
• If r = 1, then there is a perfect linear relationship that slopes up.
The Coefficient of Determination refers to how much percent Variation is around
the model. This is denoted by R². Note: If you have one you can find the other.
Simple Linear Regression is a data analysis technique that tries to find a linear
pattern in the data. In
linear regression, we use all the data to calculate a straight line which may be
used to predict the values. We will also discuss if the linear regression is
significant and if the independent variable (x) is a significant predictor of the
dependent variable (y).
The equation of line for a Simple Linear Regression (SLR) is:
�̂� = 𝛽1𝑥 + 𝛽0
Where 𝛽1 is the slope coefficient or the coefficient, 𝛽0 is the y-intercept and �̂� is
the predicted y value.
Let review our car price example. From the car price data, we also found out
what year these cars where manufactured in.
Car Price: Year
Observation 1 $ 20,000 2015
Observation 2 $ 25,000 2016
Observation 3 $ 30,000 2018
Observation 4 $ 31,000 2018
Observation 5 $ 22,500 2016
Observation 6 $ 25,000 2016
Observation 7 $ 29,500 2018
Observation 8 $ 24,000 2015
Observation 9 $ 24,500 2017
Observation 10 $ 25,000 2017
Having this information, we first want to see if there is a correlation between Year
and Car Price. Usually the older the car, cheaper the car is. As the age goes up,
the price will go down. The Price of the car depends on what Year it was
manufactured. This describes a negative correlation, but I want to see if my
assumption is correct and what the actual correlation value is.
Before we can do any calculations on the data, we will need to convert the Year to
a numeric value. Keeping the physical Year is going to skew the data and it
doesn’t make sense when we will get into analyzing and interpreting the data. If
the car was made in 2018, then this means the car will be 1 year old. 2019 – 2018
= 1. I am rounding all these to full years for ease of the example. Converting all
these Years will look like:
Car Price: Year Years
Old
Observation 1 $ 20,000 2015 4
Observation 2 $ 25,000 2016 3
Observation 3 $ 30,000 2018 1
Observation 4 $ 31,000 2018 1
Observation 5 $ 22,500 2016 3
Observation 6 $ 25,000 2016 3
Observation 7 $ 29,500 2018 1
Observation 8 $ 24,000 2015 4
Observation 9 $ 24,500 2017 2
Observation 10 $ 25,000 2017 2
Now that we have our data we can start analyzing it. To find the correlation we
will use the =CORREL( ) function in Excel.
In Excel type in the “=” and the CORREL(;put a left parentheses, then highlight the
first column; type in a comma, highlight the second column; close the
parentheses ) and hit Enter. Note: it does not matter which column you highlight
first.
Here we see that the Correlation = -.8846. This is in fact negative correlation and
agrees with our assumption. As the Age of the Car goes up, the Price of the Car
will go down. Now that we have the Correlation we can find R2.
-.8846 * – .8846 = 78.25%
Next, we will run a Regression using Excel. We will use the Data Analysis ToolPak
to run the Regression.
Go to Data – > Data Analysis
When the new window pops ups, scroll to where it says “Regression”, highlight it
and Click “OK”
Once you Click “OK”, a new window pops up
Where it will say “Input”
Input Y Range: Click in the box and highlight the y values.
Input x Range: Click in the box and highlight the x values.
Check the box that say “Labels” this will tell you that the first row has labels in it.
Output Options
Make sure the second bubble is highlighted. “New Worksheet Ply”
Residuals
Make sure you check the box for Residuals and Standardized Results
Then Click “OK”
(Remember, the x-value predicts the y-value. The Year of the Car will predict
what the Price of the Car is. This tells us that Years Old is the x-value and Price is
the y-value. This is very important to understand and remember)
It should look like this:
Once you click OK, here is the Regression Output:
Looking at the output we see the Multiple R is the correlation. We know the
Correlation is negative, but the regression will give us the positive value. Make
sure you look at the coefficients for validation. We also see that the R-squared is
78.25%. Which is what we calculated before. R-squared tells us that:
78.25% of variation in the data between Age and Price, can we have accounted
for by this model. The best R-squared value is 100%, our value is less than that,
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.884606501
R Square 0.782528661
Adjusted R Square 0.755344744
Standard Error 1725.490814
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 85706451.61 85706451.61 28.78646 0.000673381
Residual 8 23818548.39 2977318.548
Total 9 109525000
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 31959.67742 1296.435244 24.65196589 7.83E-09 28970.09239 34949.26245 28970.09239 34949.26245
Years Old -2629.032258 490.0064638 -5.365301179 0.000673 -3758.98919 -1499.075326 -3758.98919 -1499.075326
but it is still high enough to give us a good indication of what the data will look
like and it tells us that we want to interpret the model further.
Next, we want to see if this model is significant and if Years is a significant
predictor for Price. We will look at the Significance F value. Recall: if the p-value
is < alpha, then Yes this is significant.
The p-value associated with this model is .00067338.
.00067338 < .05. The p-value is in fact less than alpha. We can state that Yes,
Years is in fact a significant predictor for Price. This can be valuable information
to know if we are going out to buy a new car. Now that we know the model is
significant let’s write out the Regression Equation and interpret the values.
In the Regression Output if we look like the under Coefficients, this is where we
will find the values to write out the Regression Equation. I highlighted them in
Yellow below.
Next to those value we see the word “Intercept”, this corresponds to the y-
intercept value. And we see the words “Years Old”, this corresponds to the slope
coefficient value. Using this equation
�̂� = 𝛽1𝑥 + 𝛽0, we will write out the regression equation and replace “x” and “y”
with the actual variable names.
𝑃𝑟𝑖𝑐�̂� = −2,629.03 (𝑌𝑒𝑎𝑟𝑠 𝑂𝑙𝑑) + 31,959.68
We see that the y-intercept is $31,959.68. This means when Years Old equals 0,
the Price of a Car should be $31,959.68.
𝑃𝑟𝑖𝑐�̂� = −2,629.03 (0) + 31,959.68
𝑃𝑟𝑖𝑐�̂� = 31,959.68
This makes sense because Year 0 is 2019. So, if you bought this type of car in the
Year 2019, you will expect to pay $31,959.68.
Please note: The y-intercept while in this case does make sense does not always
have a practical meaning. The y-intercept WILL NOT make sense in every
scenario. It is OK for the y-intercept not make sense with certain problems. For
example, if you wanted to use the Weight of a Car to predict the Price, the Weight
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 31959.67742 1296.435244 24.65196589 7.83E-09 28970.09239 34949.26245 28970.09239 34949.26245
Years Old -2629.032258 490.0064638 -5.365301179 0.000673 -3758.98919 -1499.075326 -3758.98919 -1499.075326
of a Car will NEVER be 0 pounds, so the y-intercept is not meaningful and would
not have a practical meaning in the problem.
Next, we want to interpret the Slope. As the Years Old increases by 1 year, then
the Price of the Car will go down by $2,629.03. Or as the car gets older and older
the price will keep decreasing by $2,629.03 every year.
Lastly, I want to use my Regression Equation to predict prices. What would I
expect to pay for a car that was manufactured in 2014? Remember 2019 – 2014 =
5. This means the car is 5 Years Old. This is the value you want to substitute into
the Regression Equation. DO NOT put 2019 into the equation.
𝑃𝑟𝑖𝑐�̂� = −2,629.03 (𝑌𝑒𝑎𝑟𝑠 𝑂𝑙𝑑) + 31,959.68
𝑃𝑟𝑖𝑐�̂� = −2,629.03 (5) + 31,959.68
𝑃𝑟𝑖𝑐�̂� = −13,145.16 + 31,959.68
𝑃𝑟𝑖𝑐�̂� = $18,814.52
In the Year 2014, when the car is 5 Years Old, we will expect to pay $18,814.52 for
a car.
This is a good analysis for a SLR. But if we wanted to analysis the data further?
We could run a Multiple Linear Regression (MLR). Multiple Linear Regression is
just like it sounds. Instead of having only 1 x-variable, we have multiple x-
variables. In our example the x-variable was Years Old, and it did a good job at
predicting Price. But what other values could you use to predict the Price of a
Car?
One this that comes to mind is Total Miles. When you are looking to buy a car,
you also want to look at Total Miles. Usually you want a car with fewer miles on
it. The fewer miles on the car, the higher the price. Or the more miles you have a
car, the lower the price. This appears to be another negative correlation, or
relationship.
Another variable that comes to mind is a 5-star safety rating. The safer the car
the more people are willing to pay for safety. If a car has 5 stars it will be more
expensive than if a car only had 2 or 3 stars. This appears to be a positive
correlation or relationship.
You would want to run a MLR to justify and verify your claims, but these are just a
few variables you could include to turn this SLR to a MLR.
| Type | Year | Make | Model | Price | MPG(CITY) | MPG (HIGHWAY) | Weight | |||||||||||
| variable type: qualitative | variable type: quantitative | variable type: quantitative | ||||||||||||||||
| SUV | 2021 | Mazda | CX-30 | $ | 22 | 25 | 33 | 3232 lbs. | ||||||||||
| compact crossover | Toyota | rav4prime | $ | 29 | 36 | 40 | 5,530 lbs | |||||||||||
| Chrysler | Voyager | $ | 28 | 19 | 4,330 lbs. | |||||||||||||
| minivan | 2020 | kia | Sedona | $28,720 | 18 | 24 | 6,085 lbs | |||||||||||
| Dodge | grand caravan | $29,025 | 17 | 4510 lbs | ||||||||||||||
| passenger wagon | Ford | transit connect | $28,315 | 26 | 3689 lbs. | |||||||||||||
| Volkswagen | Tigwan | $25,965 | 3847 lbs. | |||||||||||||||
| 2019 | Sorento | $28,110 | 3810 lbs. | |||||||||||||||
| Honda | Odyssey | $32,110 | 4593 lbs. | |||||||||||||||
| Hyundai | Palisade | $33,665 | 4,284 lbs. |
Essay Writing Service Features
Our Experience
No matter how complex your assignment is, we can find the right professional for your specific task. Achiever Papers is an essay writing company that hires only the smartest minds to help you with your projects. Our expertise allows us to provide students with high-quality academic writing, editing & proofreading services.
Free Features
Free revision policy
$10Free bibliography & reference
$8Free title page
$8Free formatting
$8How Our Dissertation Writing Service Works
First, you will need to complete an order form. It's not difficult but, if anything is unclear, you may always chat with us so that we can guide you through it. On the order form, you will need to include some basic information concerning your order: subject, topic, number of pages, etc. We also encourage our clients to upload any relevant information or sources that will help.
Complete the order form
Once we have all the information and instructions that we need, we select the most suitable writer for your assignment. While everything seems to be clear, the writer, who has complete knowledge of the subject, may need clarification from you. It is at that point that you would receive a call or email from us.
Writer’s assignment
As soon as the writer has finished, it will be delivered both to the website and to your email address so that you will not miss it. If your deadline is close at hand, we will place a call to you to make sure that you receive the paper on time.
Completing the order and download