...The Final Project is worth 105 points. Please download this document to your computer and save it using the naming convention specified in the course syllabus. For the Final Project you will be using the MM207 Student Data Set, the survey codebook, and StatCrunch as necessary. You should enter your answers/responses directly after the question. There is no need to retype the project. After completing and saving the project, submit your project in the Final Drop Box. In the course, go to Unit 9 -> Instructor Graded Project -> StatCrunch to access the MM207 Student Data Set. When the page loads you will need to click on Data Set on the left side of the page. You do not need a StatCrunch ID or a password to access the data set; simply click the on Data Set to load the data file. Name: Crystal Card 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. 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? I usually get better grades when I spend more time doing school work so I assumed a high correlation. c) Does the number of hours spent on school work have a causal relationship with the GPA? ...
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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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