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Mm207 Final Project

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MM207 Final Project

1.
Using the MM207 Student Data Set: a) What is the correlation between student cumulative GPA and the number of hours spent on school work each week? Be sure to include the computations or StatCrunch output to support your answer. My answer :

0.27817234
(from StatCrunch):

Correlation between Q10 What is your cumulative Grade Point Average at Kaplan University? and Q11 How many hours do you spend on school work each week? is: 0.27817234

b) Is the correlation what you expected? My answer: No. I expected the correlation to be much higher because the more hours you study should equate to a much higher GPA – in theory that is.

c) Does the number of hours spent on school work have a causal relationship with the GPA? My answer: Yes. I was going to say no (because of the low correlation above), until I did a scatter plot. This shows that

MM207 Final Project there definitely is a casual relationship between study time and GPA.

yuck. There are 2 points on the right that most likely could be excluded. d) What would be the predicted GPA for a student who spends 16 hours per week on school work? Be sure to include the computations or StatCrunch output to support your prediction. My answer: 3.6

from StatCrunch Group by: Q11 How many hours do you spend on school work each week? Q11 How many hours do you spend on school work each week? 3 4 5 6 7 8 10 11

Mean 3.6666667 2 3.3775 3.0714285 3.75 3.352 2.9693334 3.6466668

n 3 1 8 7 2 5 30 3

Variance 0.33333334 NaN 0.3129357 0.42641428 0.125 0.26252 1.6706271 0.14423333

MM207 Final Project

12 13 14 15 16 hours

3.290909 4 3.93 3.7127273 3.6

11 2 2 11 3

1.4214091 0 0.0098 0.11040182 0.07

2.
Select a continuous variable that you suspect would not follow a normal distribution. a) My answer: my continuous variable is “Age” b) Create a graph for the variable you have

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...MM207 Final Project 175 total students took the survey 42+4+3+7+1+1 =58 students travel less than 5 miles to work 58/175=0.33142857142857142857142857142857 1-0.33142857142857142857142857142857=0.66857142857142857142857142857143 (1.96/.02)^2*0.33142857142857142857142857142857*0.66857142857142857142857142857143 =9604 * 0.33142857142857142857142857142857 = 3183.04 =3183.04 * 0.66857142857142857142857142857143 =2128.0896 173 answered the question 42+4+3+7+1+1 =58 students travel less than 5 miles to work 58/173=0.3352601156069364161849710982659 1-0.3352601156069364161849710982659=0.6647398843930635838150289017341 (1.96/.02)^2 * 0.6647398843930635838150289017341 = = 9604*0.3352601156069364161849710982659*0.6647398843930635838150289017341 =2140.3548397874970764141802265361 10. (-1) A professor at Kaplan University claims that the average age of all Kaplan students is 36 years old. Use a 95% confidence interval to test the professor's claim. Is the professor's claim reasonable or not? Explain. My answer: YES, it is because the intervals are roughly 36-419the professor’s claim is pretty accurate. But, 36<36.36. This is where Statistics draws the hard line, no more “roughly” or “approximately”. If the Prof had said 36.25, you would still reject because 36.25<36.36. interval are 36.36-41.137 size= 175 mean=37.94857 standard deviation=10.726628 sum=6641 6641/175=37.948571428571428571428571428571 or 37.95 sqrt of 175=13.228756555322952952508078768196 st...

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