Instructions The Final Project is worth 105 points. Please download this document to your computer and save it as something easy to find like Unit9FinalProjectYourName.

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MM207 Unit 9 Final Project

Assignment and Template

Instructions:The Final Project is worth 105 points. Please download this documentto your computer and save it as something easy to find like:Unit9FinalProjectYourName.

For thisProject you will be using the ProjectDataset in Doc Sharing under theCategory called “Final Project Assignment and Data.” This datasetis in Excel and you will use Excel to complete this Project. Youshould enter your answers/responses/graphs/charts directly after thequestion.

Aftercompleting and saving the Project, submit your Project in the DropBox. As you work on the Project, be sure to save often and also toemail it to yourself. This will avoid project-loss.

The MathCenter offers free tutoring and free project review.

Tutorials andHelp can be found in Doc Sharing, the Live Binder, and YouTubeChannel: ProfessorAmiGates (look for MM207 videos).

NOTE: ThisProject will require access to a z-table which can be found in DocSharing.

___

  1. Correlation, Scatterplots, and Prediction

  1. Which relationship is stronger, the relationship between GPA and AGE or the relationship between GPA and SAT score? Be sure to include all appropriate measures and explain and defend your answer. Is this result what you would expect, why or why not?

a). The relationship between GPA and SAT score is stronger than therelationship between and AGE and GPA. This is true because age of astudent does not determine how many marks he or she gets.Furthermore, there are younger students who have performed betterthan older students. However, if a student scored high in the GPAthen it is more likely that he or she would score higher in SATscores and vice versa. This is the result expected since the moreinput students puts in their studies is the more likely they willscore higher regardless of the age.

  1. Create and paste in a scatterplot that compares Final Exam Score and Project Score. What is the correlation (r-value)? How would you describe the correlation (positive, negative, strong, weak, medium, none)? Include the “trendline” and “equation of the trendine” as part of your scatterplot.

Thisscatter plot has a strong positive correlation with an estimated rvalue of 0.75. (r=0.75)

Theequation of the scatter plot can be written in the form

Y-Y1=m(X-X1)

{(80,80) and (55, 60)} are two points on the line.

M

80-60/80-55= 20/15

=4/3

Equation Y-60= {X-55}

Simplifiedversion: 3y=4x-40

  1. Using the equation from your above scatterplot trendline, predict (estimate) the Project score for a person who gets a final exam score of 82. Show all of your work.

Equation:Y-60={X-55}=3y=4x-40

Thereforefor final score of 82, we can find the estimated project score bysubstituting as follows.

3y=4(82)-40

3y=288

Y=96.

Final Exam Score (0 – 100)

Project Score (0 – 100)

81

45

80.872

51.49

81

46

80.872

51.75

81

53

80.872

55.86

81

56

80.872

59.55

81

60

80.872

65.27

81

61

80.872

67.00

81

67

80.872

69.00

81

69

80.872

70.53

81

71

80.872

73.00

81

75

80.872

75.00

81

75

80.872

75.71

81

78

80.872

79.00

81

81

80.872

79.00

81

82

80.872

79.35

81

83

80.872

80.00

81

85

80.872

80.00

81

86

80.872

81.00

81

87

80.872

86.00

81

87

80.872

87.00

81

88

80.872

87.25

81

89

80.872

87.38

81

89

80.872

88.75

81

89

80.872

89.00

81

89

80.872

90.00

81

90

80.872

90.00

81

91

80.872

90.00

81

92

80.872

91.00

81

93

80.872

91.50

81

94

80.872

92.46

81

96

80.872

93.99

81

97

80.872

97.23

81

97

80.872

97.25

81

98

80.872

98.67

81

99

80.872

99.66

TOTAL: 2738

2749.650

  1. MEAN = SUM OF ALL/TOTAL NUMBER

FinalExam Score = 2738/34

= 81

ProjectScore = 2749.65/34

=80.872

  1. Median scores: final exam score= 86.5

Projectscore= 83.5

  1. Mode for final score is 89 and mode for the project score is 90.

  2. Range is the difference between the highest and lowest scores.

Forthe final score = 99-45 = 54

Forproject score=99.66-51.49= 48.17

  1. Standard deviation =

Variance= (squaredsum of the deviations/N-1)

Deviations from the final exam score

Deviations from the project score

Variances final score

Variances project score

36

35.872

1296

1286.8

35

34.872

1225

1216.056

28

27.872

784

776.8484

25

24.872

625

618.6164

21

20.872

441

435.6404

20

19.872

400

394.8964

14

13.872

196

192.4324

12

11.872

144

140.9444

10

9.872

100

97.45638

6

5.872

36

34.48038

6

5.872

36

34.48038

3

2.872

9

8.248384

0

-0.128

0

0.016384

-1

-1.128

1

1.272384

-2

-2.128

4

4.528384

-4

-4.128

16

17.04038

-5

-5.128

25

26.29638

-6

-6.128

36

37.55238

-6

-6.128

36

37.55238

-7

-7.128

49

50.80838

-8

-8.128

64

66.06438

-8

-8.128

64

66.06438

-8

-8.128

64

66.06438

-8

-8.128

64

66.06438

-9

-9.128

81

83.32038

-10

-10.128

100

102.5764

-11

-11.128

121

123.8324

-12

-12.128

144

147.0884

-13

-13.128

169

172.3444

-15

-15.128

225

228.8564

-16

-16.128

256

260.1124

-16

-16.128

256

260.1124

-17

-17.128

289

293.3684

-18

-18.128

324

328.6244

7680

7676.461

Varianceof final exam score = {summed squares/n-1}

=(7680/34-1)

=232.73

StandardDeviation =

=15.255

Varianceof project score = (7676.461/34-1)

=232.62

Standarddeviation =

=15.252

  1. Comparing and Describing Data

Forthe Final Exam Score and then for the Project Score, calculate themean, median, mode, range, standard deviation, and variance. (Hint:remember to use the sample std dev and sample variance).

Final Exam Score (0 – 100)

Project Score (0 – 100)

81

45

80.872

51.49

81

46

80.872

51.75

81

53

80.872

55.86

81

56

80.872

59.55

81

60

80.872

65.27

81

61

80.872

67.00

81

67

80.872

69.00

81

69

80.872

70.53

81

71

80.872

73.00

81

75

80.872

75.00

81

75

80.872

75.71

81

78

80.872

79.00

81

81

80.872

79.00

81

82

80.872

79.35

81

83

80.872

80.00

81

85

80.872

80.00

81

86

80.872

81.00

81

87

80.872

86.00

81

87

80.872

87.00

81

88

80.872

87.25

81

89

80.872

87.38

81

89

80.872

88.75

81

89

80.872

89.00

81

89

80.872

90.00

81

90

80.872

90.00

81

91

80.872

90.00

81

92

80.872

91.00

81

93

80.872

91.50

81

94

80.872

92.46

81

96

80.872

93.99

81

97

80.872

97.23

81

97

80.872

97.25

81

98

80.872

98.67

81

99

80.872

99.66

TOTAL: 2738

2749.650

  1. MEAN = SUM OF ALL/TOTAL NUMBER

FinalExam Score = 2738/34

= 81

ProjectScore = 2749.65/34

=80.872

  1. Median scores: final exam score= 86.5

Projectscore= 83.5

  1. Mode for final score is 89 and mode for the project score is 90.

  2. Range is the difference between the highest and lowest scores.

Forthe final score = 99-45 = 54

Forproject score=99.66-51.49= 48.17

  1. Standard deviation =

Variance= (squaredsum of the deviations/N-1)

Deviations from the final exam score

Deviations from the project score

Variances final score

Variances project score

36

35.872

1296

1286.8

35

34.872

1225

1216.056

28

27.872

784

776.8484

25

24.872

625

618.6164

21

20.872

441

435.6404

20

19.872

400

394.8964

14

13.872

196

192.4324

12

11.872

144

140.9444

10

9.872

100

97.45638

6

5.872

36

34.48038

6

5.872

36

34.48038

3

2.872

9

8.248384

0

-0.128

0

0.016384

-1

-1.128

1

1.272384

-2

-2.128

4

4.528384

-4

-4.128

16

17.04038

-5

-5.128

25

26.29638

-6

-6.128

36

37.55238

-6

-6.128

36

37.55238

-7

-7.128

49

50.80838

-8

-8.128

64

66.06438

-8

-8.128

64

66.06438

-8

-8.128

64

66.06438

-8

-8.128

64

66.06438

-9

-9.128

81

83.32038

-10

-10.128

100

102.5764

-11

-11.128

121

123.8324

-12

-12.128

144

147.0884

-13

-13.128

169

172.3444

-15

-15.128

225

228.8564

-16

-16.128

256

260.1124

-16

-16.128

256

260.1124

-17

-17.128

289

293.3684

-18

-18.128

324

328.6244

7680

7676.461

Varianceof final exam score = {summed squares/n-1}

=(7680/34-1)

=232.73

StandardDeviation =

=15.255

Varianceof project score = (7676.461/34-1)

=232.62

Standarddeviation =

=15.252

  1. Which two numerical measures offer you the best information for comparing performance between these two assignments? Use these two numerical measures to describe and compare student performance between the final exam and the Project.

Thetwo numerical measures that offer the best information for comparingperformance between final exams scores and the project scores are theSAT scores and the amount of hours per week spent on studying. Thehigher the student scored on the SATs corresponded to the high scoresin the exam and project scores. The amount of hours spent studyingper week is also directly proportional to the performance of thestudents on the two assignments.

  1. Which variable, Final Exam Score or Project Score has greater variation of data? Which two numerical measures are best to offer this information? Choose the two numerical measures and use them to describe and compare the variation between the two variables. What does the variation tell you about student performance?

Thefinal exam scores and project scores have an almost similar variationof data. The amount of hours spent per week studying, the SAT scoresand the GPA scores can best describe the similarity in variations.The higher a student’s GPA scores are, the higher the scores in thefinal exams and the project. The section of class and the year inschool are also significant in the variation because if a student inthe 5thyear, section 4, he or she would need to work extra hard to get goodgrades because it is the final year.

  1. Who did better on the Final Exam, males or females? Justify your answer using at least THREE (3) numerical measures AND a bar graph. (Hint: Your bar graph will only show the mean and median values for the males and females).

Themales performed better than the females in the final exam. This isbecause the males had a MEAN score of 82.588 while the females had aMEAN score of 78.47. Despite the males’ performing better, theyrecorded the minimum score of 45 while females recorded the maximumscore of 99. The males’ performed better because all of them scoredon and above the 75thpercentile except for two males. Only ten females managed to scoreabove the 75thpercentile.

AGraph of Performance in the final Exams against males and females.

  1. Relative and Absolute Differences

  1. Sally and Ron have decided to go back to school. Sally is 28 and Ron is 25. Based on their relative position, which of them (Sally or Ron) would be farther away from the average age of their gender group? Show all steps and work. (Hint: You will need to calculate z-scores as part of this question, which are the number of standard deviations from the mean).

Meanage of females =27

Meanage of males =26

Standarddeviation: square root of the variance.

ForRon: mean age is 26

Variancefor males = 152/17-1

=9.5

Standarddeviation =

=3.08

=(152-26)/3.08

=40.91

Variancefor females = 288/17-1

=18

Standarddeviation =

=4.24

Zscore = (288-18)/4.24

63.68

Fromthe calculations, (40.91-25) =15.91 and (63.68-28) =35.68, Sallywould be farther away from the

  1. Sally has a GPA of 3.35. What percentage of the students have a GPA above her GPA? Show all work. (Note: GPA is normally distributed).

MeanGPA score = 3.27

Variance= 0.125516

S.D= square root of the variance

0.3543

Thenormal distribution function is given by:

Thenormal distribution is 0.4107 therefore the percentage of thestudents above her GPA is 0.4107*100 = 41%

  1. SAT scores are normally distributed and can range from 0 to 1600. Ron’s SAT score is 800. What percentage of the MALE Student SAT scores in the dataset are below Ron’s SAT score? What would you say about Ron’s performance on the SAT test as compared to the other male students in the dataset? Show all work. (Hint: use z scores and the z table to solve this).

Themean SAT scores for the males is = 1092

Thevariance is =1153089/(17-1)

=1076

Standarddeviation = (square root of variance)

=32.80

Thenormal distribution function is given by:

P&nbsp(&nbsp0−1092/32.8&ltXμσ&lt800−1092/32.8)

Thepercentage of the male population that scored below 800 in the SATscores is 0. This means that Ron’s performance was quite poor ascompared to the other males’ performance on the SAT scores.

  1. Frequencies and Graphs

  1. A person’s GPA can be categorized into one of five classifications:

F:0.00 – 0.49

D:0.50 – 1.49

C:1.50 – 2.49

B:2.50 – 3.49

A:3.50 – 4.00

Usethese 5 categories and create a relative and cumulative frequencychart for all the GPAs in the dataset. You can do this by hand or youcan use Excel.

Arelative frequency chart showing the GPA performance

GPA SCORES

FREQUENCY

2.59

2.33

3.49

3.14

3.78

3.40

3.91

3.52

3.47

3.12

3.36

3.02

3.14

2.83

2.80

2.52

2.93

2.64

3.49

3.14

3.32

2.99

3.72

3.35

3.08

2.77

3.90

3.51

3.33

3.00

1.78

1.60

3.22

2.90

3.36

3.02

3.47

3.12

2.89

2.60

3.34

3.01

2.96

2.66

3.10

2.79

3.38

3.04

3.09

2.78

3.59

3.23

3.93

3.54

3.02

2.72

3.45

3.11

3.49

3.14

3.54

3.19

3.49

3.14

3.82

3.44

1.86

1.67

TOTAL:

ACumulative Frequency Chart

Range

Frequency

Cumulative frequency

F: 0.00 – 0.49

0

0

D: 0.50 – 1.49

0

0

C: 1.50 – 2.49

2

2

B: 2.50 – 3.49

24

26

A: 3.50 – 4.00

8

34

APie Chart showing the relative frequencies of the GPA scores

  1. Create a bar chart to show the frequencies of each letter grade.

.

  1. Confidence Intervals: The dataset for this Project represents a sample of data from a larger population.

  1. Use this dataset sample to calculate the 95% confidence interval for the true population mean time that all students spend studying per week. If a student spends 15 hours per week studying, is this significantly different from the population mean? Explain. Show all work and any use of Excel. You may choose to use Excel or you may do this by hand.

Averagehours spend on studying

Hours

1.6

2.4

3.5

4.6

5.4

6.0

6.0

6.3

6.5

6.7

6.9

7.4

8.1

8.3

8.7

8.9

9.0

9.0

10.6

10.9

11.0

11.3

11.4

11.6

11.9

12.3

12.4

12.8

13.3

13.5

13.7

14.2

17.2

Average = 9.5

Confidenceinterval = 95%

Meanhours per week spent studying= 9.5 hours

Populationsize = 34

Samplesize= 1

Thereforethe confidence interval is 42.72. if a student spends 15 hoursstudying per, it is relatively higher than the population mean whichis 9.5 hours.

  1. In your own words, explain why and how sample statistics are used to estimate population parameters. Using the dataset, create an example to support your explanation.

Samplestatistics are very useful in estimating population parameters. Thisis because, in most cases, the population to be studied may be toolarge or have an infinite number. The best way to go about it is totake a small section of the population and study it.

Forinstance from the data set, we have selected a small group of 34students, represented by an equal number of males and females from avast population of university students. Dealing with statistics fromthe whole population would prove tedious and time consuming.