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Optional Midterm Project

Optional Midterm Project

Directions:

1. You will have almost one weeks to complete this examination.

2. You may use your notes, homework examples, code files from class, your textbook, other statistical

eferences, and your personal knowledge. You may NOT consult other students or faculty members.

3. If you have questions or need clarification, contact me by email at XXXXXXXXXX.

Please type the following pledge at the beginning of your na

ative document and type your name after it as

your signature if you comply with this pledge. If you do not include this pledge, your exam may be returned

to you ungraded.

On my honor as a student, I have neither given nor received aid on this examination outside

the scope of the directions.

To Complete this assignment you would submit one or two on documents on Canvas.

Either:

1. A R code script with the code used to prepare variables, tables, and graphs. (.R)

2. A text or word processing document with the na

ative and any relevant tables and graphs (e.g. .docx,

.pdf, .html). This document should stand alone, meaning that all relevant statistics and other information

should be included in tables or graphs within this document.

Or:

1. An html file with the R code and text write up in the same document. This document should include

all the content mentioned above.

DO NOT TURN IN A Lname_midterm.Rmd FILE

For each document use the following naming convention,

Lname_midterm.suffix

where Lname is your capitalized last name, and suffix is the relevant suffix (.R, .html, .docx). For example,

if I were turning in homework, I could include the following files:

Mu

ah_midterm.R

Mu

ah_midterm.docx

O

Mu

ah_midterm.html

NOTE: the text or Word document should be self-contained. That means all the output

needed for the arguments made must also be in this document. It is NOT sufficient to refe

to the R code script.

1

XXXXXXXXXX

I. (100 points)

Import the mo

ison2.csv file for this question.

This study is aimed at understanding which characteristics of a course instructor are important to students.

Using course instructor evaluations we sought to determine if clarity of presentation and instructor content

knowledge are important in predicting how well students’ rate the instructor. The data contain information

elated to student evaluations of instructors of Master’s of Business (MBA) courses. The goal is to estimate

the effect of clarity and knowledge as predictors of the overall instructor evaluation. The outcome is instructo

course evaluation of an MBA course, with two predictors being clarity and knowledge.

The definition of the variables in the data follow.

Variable name Variable Label

insteval Course evaluation of instructo

clarity Clarity of presentations

knowledg Instructors content knowledge

Guiding questions:

The following questions should guide your write up. BUT THE WRITE UP SHOULD BE FORMAT-

TED AS A VERY ABBREVIATED MANUSCRIPT. It should look like the examples given in class.

Do not include the question followed by the answers.

a. Generate a concise table with descriptive statistics for the important variables in the data set.

. Generate a concise co

elation table for the important variables in the data set.

c. Determine the regression model (equation) in raw score form. Include the estimated equation for each

and provide your interpretation for the meaning of the regression coefficients.

d. Evaluate the overall fit of the model using the multiple R2 and the co

esponding statistical test, and

explain the meaning of this test. Explain the R2 as an effect size measure.

e. Determine if both predictors are meaningful in the regression model. Be sure to discuss effect size and

tests of the regression coefficients. Explain your results.

f. Check the data for assumptions, outliers, and influential cases. Write a

ief paragraph summarizing

what you found.

g. Interpret these results and how they are related to the original research question.

2

Directions:

I. (100 points)

Guiding questions:

Midterm Ru

ic (2)

Criteria

Ratings

Pts

This criterion is linked to a Learning OutcomeQ1.a Descriptives

5 pts

Full Marks

0 pts

No Marks

5 pts

This criterion is linked to a Learning OutcomeQ1.b Co

elations

5 pts

Full Marks

0 pts

No Marks

5 pts

This criterion is linked to a Learning OutcomeQ1.c Equations

10 pts

Full Marks

0 pts

No Marks

10 pts

This criterion is linked to a Learning OutcomeQ1.d R squared significance

5 pts

Full Marks

0 pts

No Marks

5 pts

This criterion is linked to a Learning OutcomeQ1.e predictors meaning

10 pts

Full Marks

0 pts

No Marks

10 pts

This criterion is linked to a Learning OutcomeQ1.f Iterpret coefficients

5 pts

Full Marks

0 pts

No Marks

5 pts

This criterion is linked to a Learning OutcomeQ2.a.1. Descriptive statistics

5 pts

Full Marks

0 pts

No Marks

5 pts

This criterion is linked to a Learning OutcomeQ2.a.2. Co

elation Table

5 pts

Full Marks

0 pts

No Marks

5 pts

This criterion is linked to a Learning OutcomeQ2.b Regression Equation

10 pts

Full Marks

0 pts

No Marks

10 pts

This criterion is linked to a Learning OutcomeQ2b.2. Adequacy of equation

10 pts

Full Marks

0 pts

No Marks

10 pts

This criterion is linked to a Learning OutcomeQ2.c.1. Determine predictor meaningfulness.

10 pts

Full Marks

0 pts

No Marks

10 pts

This criterion is linked to a Learning OutcomeQ2.c.2. Describe results in APA style

10 pts

Full Marks

0 pts

No Marks

10 pts

This criterion is linked to a Learning OutcomeQ2.d. Interpret results

10 pts

Full Marks

0 pts

No Marks

10 pts

Total Points: 100

Midterm Ru

ic (2)

Answered Same DayApr 14, 2022

Solution a:

Code & Output:

summary(data)

insteval clarity knowledg

Min. :1.000 Min. :1.000 Min. :1.000

1st Qu.:2.000 1st Qu.:2.000 1st Qu.:1.000

Median :2.000 Median :3.000 Median :1.000

Mean :2.406 Mean :2.844 Mean :1.438

3rd Qu.:3.000 3rd Qu.:4.000 3rd Qu.:2.000

Max. :4.000 Max. :5.000 Max. :3.000

Interpretation:

The minimum and maximum for “insteval” variable is 1 and 4. The mean and median for this variable is obtained as 2.406 and 2 respectively. The first and third quarter for the variable is 2 and 3 respectively.

The minimum and maximum for “clarity” variable is 1 and 5. The mean and median for this variable is obtained as 2.844 and 3 respectively. The first and third quarter for the variable is 2 and 4 respectively.

The minimum and maximum for “knowledg” variable is 1 and 3. The mean and median for this variable is obtained as 1.438 and 1 respectively. The first and third quarter for the variable is 1 and 2 respectively.

Solution b:

Code & Output:

cor(data)

insteval clarity knowledg

insteval 1.0000000 0.86180913 0.28182052

clarity 0.8618091 1.00000000 0.05725983

knowledg 0.2818205 0.05725983 1.00000000

Interpretation:

From the co

elation table, it is obtained that the co

elation between clarity and insteval variable is 0.861 which implies there is very strong positive co

elation between clarity and insteval. The co

elation between insteval and knowledg is 0.282 which implies there is poor positive association between these two variables. The co

elation between clarity and knowledge is 0.057 which implies there is very poor positive co

elation between clarity and knowledge.

Solution c:

Code & Output:

fit=lm(insteval~.,data=data)

summary(fit)

Call:

lm(formula = insteval ~ ., data = data)

Residuals:

Min 1Q Median 3Q Max

-0.74654 -0.31129 0.00139 0.25346 0.70083

Coefficients:

Estimate Std. E

or t value Pr(>|t|)

(Intercept) 0.19391 0.23712 0.818 0.42016

clarity 0.62604 0.06184 10.123 5e-11 ***

knowledg 0.30055 0.10801 2.783 0.00938 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard e

or: 0.3716 on 29 degrees of freedom

Multiple R-squared: 0.7969, Adjusted R-squared: 0.7829

F-statistic: 56.91 on 2 and 29 DF, p-value: 9.134e-11

Interpretation:

From the regression analysis, the regression equation for predicting is obtained as below:

From the equation the coefficient for clarity is obtained as 0.62604 and coefficient for knowledge is obtained for 0.30055 which implies for one unit change in clarity, keeping others constant, insteval will increase by 0.19391 + 0.62604 unit and one unit change in knowledge , keeping others constant, insteval will increase by 0.19391 + 0.30055...

Code & Output:

summary(data)

insteval clarity knowledg

Min. :1.000 Min. :1.000 Min. :1.000

1st Qu.:2.000 1st Qu.:2.000 1st Qu.:1.000

Median :2.000 Median :3.000 Median :1.000

Mean :2.406 Mean :2.844 Mean :1.438

3rd Qu.:3.000 3rd Qu.:4.000 3rd Qu.:2.000

Max. :4.000 Max. :5.000 Max. :3.000

Interpretation:

The minimum and maximum for “insteval” variable is 1 and 4. The mean and median for this variable is obtained as 2.406 and 2 respectively. The first and third quarter for the variable is 2 and 3 respectively.

The minimum and maximum for “clarity” variable is 1 and 5. The mean and median for this variable is obtained as 2.844 and 3 respectively. The first and third quarter for the variable is 2 and 4 respectively.

The minimum and maximum for “knowledg” variable is 1 and 3. The mean and median for this variable is obtained as 1.438 and 1 respectively. The first and third quarter for the variable is 1 and 2 respectively.

Solution b:

Code & Output:

cor(data)

insteval clarity knowledg

insteval 1.0000000 0.86180913 0.28182052

clarity 0.8618091 1.00000000 0.05725983

knowledg 0.2818205 0.05725983 1.00000000

Interpretation:

From the co

elation table, it is obtained that the co

elation between clarity and insteval variable is 0.861 which implies there is very strong positive co

elation between clarity and insteval. The co

elation between insteval and knowledg is 0.282 which implies there is poor positive association between these two variables. The co

elation between clarity and knowledge is 0.057 which implies there is very poor positive co

elation between clarity and knowledge.

Solution c:

Code & Output:

fit=lm(insteval~.,data=data)

summary(fit)

Call:

lm(formula = insteval ~ ., data = data)

Residuals:

Min 1Q Median 3Q Max

-0.74654 -0.31129 0.00139 0.25346 0.70083

Coefficients:

Estimate Std. E

or t value Pr(>|t|)

(Intercept) 0.19391 0.23712 0.818 0.42016

clarity 0.62604 0.06184 10.123 5e-11 ***

knowledg 0.30055 0.10801 2.783 0.00938 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard e

or: 0.3716 on 29 degrees of freedom

Multiple R-squared: 0.7969, Adjusted R-squared: 0.7829

F-statistic: 56.91 on 2 and 29 DF, p-value: 9.134e-11

Interpretation:

From the regression analysis, the regression equation for predicting is obtained as below:

From the equation the coefficient for clarity is obtained as 0.62604 and coefficient for knowledge is obtained for 0.30055 which implies for one unit change in clarity, keeping others constant, insteval will increase by 0.19391 + 0.62604 unit and one unit change in knowledge , keeping others constant, insteval will increase by 0.19391 + 0.30055...

SOLUTION.PDF## Answer To This Question Is Available To Download

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