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Preparing for the
GM 533 Final Exam
 Professor Brent Heard
GM 533 Final Exam Prep
• Summary of topics covered
    – Week 1 - Descriptive Statistics: includes central tendency, dispersion, and the
      shape of the distribution, in numbers, pictures, and tables.
    – Week 2 – Probability: includes 3 major problem types, and their most
      important variations: contingency tables, expected value, and the binomial
      distribution.
    – Week 3 – Probability continued: includes the normal distribution, its
      application to sampling distributions, and its most important variations.
    – Week 4 – Confidence intervals and sample size determinations, and their most
      important variations.
    – Week 5 – Hypothesis testing: includes the 5-step hypothesis testing
      procedure, applied to means and proportions, and its most important
      variations.
    – Week 6 – Simple linear regression: includes interpreting Minitab output for
      point estimates, hypothesis tests, and confidence intervals.
    – Week 7 – Multiple regression: includes the same elements as simple
      regression, but also includes the application to multiple independent
      variables.
• Examples and topic areas follow
GM 533 Final Exam Prep
•   Sample Question on Binomial Distribution
     – Assume that a study was done finding that 70 percent of males in Georgia are
       football fans. If a researcher asks 8 Georgia Males if they are fans, the
       following binomial distribution would be applicable. What is the probability
       that at least 5 will be football fans?


                                    n           p
                                    8          0.7

                            x      P( x)    Cumulative
                            0     0.0001      0.0001
                            1     0.0012      0.0013
                            2     0.0100      0.0113
                            3     0.0467      0.0580
                            4     0.1361      0.1941
                            5     0.2541      0.4482
                            6     0.2965      0.7447
                            7     0.1977      0.9424
                            8     0.0576      1.0000
GM 533 Final Exam Prep
•   Sample Question on Binomial Distribution
     – Assume that a study was done finding that 70 percent of males in Georgia are
       football fans. If a researcher asks 8 Georgia Males if they are fans, the
       following binomial distribution would be applicable. What is the probability
       that at least 5 will be football fans?


             n            p
             8           0.7
                                         “At least 5” is the probability that 5, 6, 7
                                         or 8 will be fans. Simply add those
     x      P( x)    Cumulative          probabilities.
     0     0.0001      0.0001
     1     0.0012      0.0013
     2     0.0100      0.0113                    0.2541
     3     0.0467      0.0580                    0.2965
     4     0.1361      0.1941                    0.1977
                                                                     My total is 0.8059
     5     0.2541      0.4482                    0.0576              or about 81% which
     6     0.2965      0.7447                                        is the probability of
     7     0.1977      0.9424      SUM           0.8059
                                                                     at least 5 being
     8     0.0576      1.0000
                                                                     football fans.
GM 533 Final Exam Prep
• Analysis Example
       – 9 members of the local college baseball team had the following number for
         extra base hits for the year. Using the Minitab output given, determine:
          A. Mean
          B. Standard Deviation
          C. Range
          D. Median
          E.   The range of the data that would contain 68% of the results.
Data
7
9
4
24
15
17
15
6
29
       –   Minitab Follows
GM 533 Final Exam Prep
Descriptive Statistics: Extra Base Hits

Variable     N N* Mean SE Mean StDev Minimum Q1 Median      Q3
Extra Base Hits 9 0 14.00 2.82 8.47 4.00 6.50 15.00 20.50

Variable     Maximum
Extra Base Hits 29.00
                                                                    This is technically NOT
Stem-and-Leaf Display: Extra Base Hits                              the correct Median.
Stem-and-leaf of Extra Base Hits N = 9    A.   Mean
Leaf Unit = 1.0                           B.   Standard Deviation
                                          C.   Range (Max – Min = 29 – 4 = 25)
                                          D.    Median (I would enter data into Excel to
 4 0 4679
                                               find the Median is 15 or do by ordering
(3) 1 557
 2 2 49                                        data and identifying)
                                          E.   The range of the data that would contain
                                               68% of the results. (Mean – Std Dev,
                                               Mean + Std Dev) which is (14 – 8.47, 14 +
                                               8.47) or (5.53,22.47)
GM 533 Final Exam Prep



          Here I used

          =MEDIAN(A1:A9)

          It returned the median of 15.
GM 533 Final Exam Prep
• Sample Question on Hypothesis Testing
  – Pepito’s Pizza Works is putting pizzas out by delivery as fast as they
    can. Pepito’s claims they can deliver pizzas within their delivery area in
    less than 29 minutes. You are given the following data from a sample.
    Sample size: 120 Deliveries
    Population standard deviation: 1.4
    Sample mean: 28.3
    Formulate a hypothesis test to evaluate the claim.
GM 533 Final Exam Prep
• Sample Question on Hypothesis Testing
  – Pepito’s Pizza Work is putting pizzas out by delivery as fast as they can.
    Pepito’s claims they can deliver pizzas within their delivery area in less
    than 29 minutes. You are given the following data from a sample.
    Sample size: 120 Deliveries
    Population standard deviation: 1.4
    Sample mean: 28.3
    Formulate a hypothesis test to evaluate the claim.

  –   Answer: Ho: µ ≥ 29, Ha : µ < 29
  –   (In this case, the claim was Ha)
  –   Remember Ho always contains equality (It will either be =, ≤ or ≥)
  –   Ha will be either ≠, < or >
GM 533 Final Exam Prep
• Confidence interval Example
  – Acme computers needs to find a new vendor for
    their hard drives. They are considering using
    Howie’s Hard Drives as a vendor. Acme’s
    requirement is that 95% of the hard drives last
    24000 hours ± 2000 hours. The following data is
    from an independent source who evaluated
    Howie’s. Should Acme buy from Howie’s? Explain
    your answer. (Follows on next page)
GM 533 Final Exam Prep
•   Mean = 24500
•   Sample Standard Deviation 2250
•   Min 21402
•   Max 29463
•   Margin of Error 4500

• Answer Follows
GM 533 Final Exam Prep
• No, Acme shouldn’t buy from Howie’s looking
  at their requirements (24000 – 2000, 24000 +
  2000) which is (22000, 26,000). Based on the
  results given, Howie’s would yield a tolerance
  of (24500 – 2*2250, 24500+2*2250) which is
  (20000, 29000). This does not meet Acme’s
  requirement. You could also see this by
  looking at the margin of error.
GM 533 Final Exam Prep
• Example on Pivot/Contingency Tables
• The table below gives the number of cars of
  various colors and the state tag on the car for
  a parking lot in a mall close to DC.
                     VA   MD   DC Other State Total
             Blue     4    8    9     3        24
             Black    9    7   11     8        35
             White   12   14   21     15       62
             Other   37   10   29     35       111
             Total   62   39   70     61      232
GM 533 Final Exam Prep
• Based on the table, find the probability that a
  car is from VA or MD.
• Based on the table, given that a car is from
  DC, find the probability it is black.
GM 533 Final Exam Prep


        VA   MD        DC Other State Total   Find the probability that a car is
Blue     4    8         9     3        24     from VA or MD. Add 62 + 39 to get
Black    9    7        11     8        35     So the answer would be 91/232 or it’s
White   12   14        21     15       62     decimal form.
Other   37   10        29     35       111
Total   62   39        70     61      232
        62 + 39 = 91
GM 533 Final Exam Prep


        VA   MD   DC Other State Total
                                         Given that a car is from DC, find
Blue     4    8    9     3        24
                                         the probability it is black.
Black    9    7   11     8        35
                                         Given that it is from DC means we are only
White   12   14   21     15       62     dealing with the 70 cars from DC.
Other   37   10   29     35       111    There are 11 of those that are black, so the
Total   62   39   70     61      232     probability is 11/70
GM 533 Final Exam Prep
• Normal Distribution Example
  – The number of students who use the dining hall at
    an urban college on a given day is normally
    distributed with a mean of 1578 students and a
    standard deviation of 274 students.
GM 533 Final Exam Prep
• I’m suggesting Excel to work these types of
  problems even if you are given partial Minitab
  results.
• Go to
http://highered.mcgraw-
  hill.com/sites/0070620164/student_view0/exc
  el_templates.html
And download the template titled Normal
  Distribution.
GM 533 Final Exam Prep
• It will look something like this when you open
  it
GM 533 Final Exam Prep
• Before doing anything else, click the “Review” tab
  at the top of Excel (Between Data and View), then
  click “Unprotect Sheet”.

                                         After clicking
                                         “Unprotect
                                         Sheet” it will say
                                         “Protect Sheet.”
                                         Leave it that way
                                         and save to your
                                         computer. You
                                         now have a cool
                                         Normal
                                         Distribution
                                         Calculator.
GM 533 Final Exam Prep
• Back to our problem….
• Questions
• What is the probability that less than 1400
  students will use the dining hall?
• What is the probability that more than 1700 will
  use the dining hall?
• What is the probability that between 1400 and
  1600 students will use the dining hall?
• Get that Normal Distribution Excel Calculator
  ready and be amazed!
GM 533 Final Exam Prep
GM 533 Final Exam Prep
GM 533 Final Exam Prep
                                       This gives
This gives                             you area to
you area to                            the right
the left                               based on
based on                               your mean
your mean                              and
and                                    standard
standard                               deviation.
deviation.
                                   This gives
                                   you area
                                   between
                                   two values
                                   based on
                                   your mean
                                   and
                                   standard
                                   deviation.
GM 533 Final Exam Prep
                                                           I entered
I entered                                                  1700 in the
1400 in the                                                green cell
green cell                                                 which gives
which gives                                                me the
me the                                                     probability
probability                                                of more
of less than                                               than 1700
1400                                                       students
students                                                   using the
using the                                                  dining hall.
dining hall.                                               The answer
The answer                                                 is 0.3281
is 0.2580            I entered 1400 in the left cell and
                     1600 in the right green cell which
                     gives me the probability of between
                     1400 and 1600 students using the
                     dining hall. The answer is 0.2740
GM 533 Final Exam Prep
• Another Confidence Interval Example
   – I randomly sampled 18 engineers where I work and asked
     them how many projects they have worked on in the last
     five years. The sample mean was 21, with a standard
     deviation of 5. What is the mean number of projects of all
     engineers at my research center? Why? What is the 95%
     confidence interval for the population mean? You are
     given the information below from Minitab.

One-Sample T
N Mean StDev SE Mean 95% CI
18 21.00 5.00 1.18  (18.51, 23.49)
GM 533 Final Exam Prep
• Another Confidence Interval Example
   – I randomly sampled 18 engineers where I work and asked them
     how many projects they have worked on in the last five years.
     The sample mean was 21, with a standard deviation of 5. What
     is the mean number of projects of all engineers at my research
     center? Why? What is the 95% confidence interval for the
     population mean? You are given the information below from
     Minitab.

Answer:
21 projects would be the best estimate for the mean. I
  would expect 95% of the population mean to fall between
  18.51 and 23.49 projects. The t is used because of the
  sample size.
GM 533 Final Exam Prep
• Regression Example
  – I did an analysis to determine if the number of
    hours studied for a final exam related to the Final
    Exam grade for students. On the sheets that
    follow you will see what my Minitab results were.
General Regression Analysis: Final Grade versus Hours of   Analysis of Variance
Study
                                                           Source       DF Seq SS Adj SS Adj MS     F     P
Regression Equation                                        Regression      1 8090.71 8090.71 8090.71 171.274
                                                           0.000000
Final Grade = 34.2845 + 1.45508 Hours of Study              Hours of Study 1 8090.71 8090.71 8090.71 171.274
                                                           0.000000
                                                           Error       22 1039.25 1039.25 47.24
Coefficients                                                Lack-of-Fit 18 936.58 936.58 52.03 2.027 0.259215
                                                            Pure Error    4 102.67 102.67 25.67
Term         Coef SE Coef   T P                            Total       23 9129.96
Constant     34.2845 3.38091 10.1406 0.000
Hours of Study 1.4551 0.11118 13.0872 0.000
                                                           Fits and Diagnostics for Unusual Observations
Summary of Model                                             Final
                                                           Obs Grade Fit SE Fit Residual St Resid
S = 6.87303 R-Sq = 88.62%       R-Sq(adj) = 88.10%         24 19 37.1947 3.17993 -18.1947 -2.98609 R
PRESS = 1426.21 R-Sq(pred) = 84.38%
                                                           R denotes an observation with a large standardized residual.
GM 533 Final Exam Prep

Predicted Values for New Observations

New Obs Fit SE Fit    95% CI        95% PI
  1 48.8353 2.41382 (43.8294, 53.8413) (33.7280, 63.9426)


Values of Predictors for New Observations

        Hours
         of
New Obs Study
  1 10
GM 533 Final Exam Prep
I did an analysis to determine if the number of hours studied for a final exam related to the
    Final Exam grade for students. On the sheets that follow you will see what my Minitab
    results were.
Answer the following questions.
Determine the regression equation.

What conclusions are possible using the meaning of bo (intercept) and b1 (regression coefficient) in this
   problem?

What does the coefficient of determination (r-squared) mean?


Calculate the coefficient of correlation and explain what it means.

Does this data provide significant evidence (a=0.05) that the final exam grade is associated with the hours
   studied? Find the p-value and interpret.

Determine the predicted grade for someone who spends 10 hours studying for the final exam.

What is the 95% confidence interval for the score for spending 10 hours studying on the test? What conclusion is
   possible using this interval?
GM 533 Final Exam Prep
I did an analysis to determine if the number of hours studied for a
   final exam related to the Final Exam grade for students. On the
   sheets that follow you will see what my Minitab results were.
Answer the following questions.

Determine the regression equation. y= 34.2845 + 1.45508x


What conclusions are possible using the meaning of bo (intercept) and b1
  (regression coefficient) in this problem? For each hour of study the final grade
  is increased by about 1.5 points (1.45508). bo represents the y intercept or
   34.2845 in our case. It is the score that a student could expect to get without
   studying.

What does the coefficient of determination (r-squared) mean? The .886 means
  that 88.6 percent of the variability of the final grade can be explained by the
  number of study hours. The other 11.4% would be due to something else or
  be unexplained.
GM 533 Final Exam Prep

Calculate the coefficient of correlation and explain what it means. Square Root of
(0.886) is 0.942 which is r, the correlation coefficient. With a value this close to
one, we could say there is strong positive correlation.

Does this data provide significant evidence (a=0.05) that the final exam grade is
associated with the hours studied? Find the p-value and interpret. Yes, the p value
was 0. If it were above 0.05, I would have said “no.”

Determine the predicted grade for someone who spends 10 hours studying for the
final exam. 48.8353

What is the 95% confidence interval for the score for spending 10 hours studying
on the test? What conclusion is possible using this interval? (43.8294, 53.8413)
We would be 95% confident that if someone studied for 10 hours they would
score on average between those two values.
GM 533 Final Exam Prep
• Multiple Regression
  – Be able to identify the multiple regression output
    from a Minitab analysis
     •   Equation
     •   F value
     •   p value
     •   Confidence intervals
     •   (Basically reading from Minitab output)
GM 533 Final Exam Prep
• I will post these charts in the “Stat Cave” at
  – www.facebook.com/statcave

  – Good Luck!

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Gm533 finalprep

  • 1. Preparing for the GM 533 Final Exam Professor Brent Heard
  • 2. GM 533 Final Exam Prep • Summary of topics covered – Week 1 - Descriptive Statistics: includes central tendency, dispersion, and the shape of the distribution, in numbers, pictures, and tables. – Week 2 – Probability: includes 3 major problem types, and their most important variations: contingency tables, expected value, and the binomial distribution. – Week 3 – Probability continued: includes the normal distribution, its application to sampling distributions, and its most important variations. – Week 4 – Confidence intervals and sample size determinations, and their most important variations. – Week 5 – Hypothesis testing: includes the 5-step hypothesis testing procedure, applied to means and proportions, and its most important variations. – Week 6 – Simple linear regression: includes interpreting Minitab output for point estimates, hypothesis tests, and confidence intervals. – Week 7 – Multiple regression: includes the same elements as simple regression, but also includes the application to multiple independent variables. • Examples and topic areas follow
  • 3. GM 533 Final Exam Prep • Sample Question on Binomial Distribution – Assume that a study was done finding that 70 percent of males in Georgia are football fans. If a researcher asks 8 Georgia Males if they are fans, the following binomial distribution would be applicable. What is the probability that at least 5 will be football fans? n p 8 0.7 x P( x) Cumulative 0 0.0001 0.0001 1 0.0012 0.0013 2 0.0100 0.0113 3 0.0467 0.0580 4 0.1361 0.1941 5 0.2541 0.4482 6 0.2965 0.7447 7 0.1977 0.9424 8 0.0576 1.0000
  • 4. GM 533 Final Exam Prep • Sample Question on Binomial Distribution – Assume that a study was done finding that 70 percent of males in Georgia are football fans. If a researcher asks 8 Georgia Males if they are fans, the following binomial distribution would be applicable. What is the probability that at least 5 will be football fans? n p 8 0.7 “At least 5” is the probability that 5, 6, 7 or 8 will be fans. Simply add those x P( x) Cumulative probabilities. 0 0.0001 0.0001 1 0.0012 0.0013 2 0.0100 0.0113 0.2541 3 0.0467 0.0580 0.2965 4 0.1361 0.1941 0.1977 My total is 0.8059 5 0.2541 0.4482 0.0576 or about 81% which 6 0.2965 0.7447 is the probability of 7 0.1977 0.9424 SUM 0.8059 at least 5 being 8 0.0576 1.0000 football fans.
  • 5. GM 533 Final Exam Prep • Analysis Example – 9 members of the local college baseball team had the following number for extra base hits for the year. Using the Minitab output given, determine: A. Mean B. Standard Deviation C. Range D. Median E. The range of the data that would contain 68% of the results. Data 7 9 4 24 15 17 15 6 29 – Minitab Follows
  • 6. GM 533 Final Exam Prep Descriptive Statistics: Extra Base Hits Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Extra Base Hits 9 0 14.00 2.82 8.47 4.00 6.50 15.00 20.50 Variable Maximum Extra Base Hits 29.00 This is technically NOT Stem-and-Leaf Display: Extra Base Hits the correct Median. Stem-and-leaf of Extra Base Hits N = 9 A. Mean Leaf Unit = 1.0 B. Standard Deviation C. Range (Max – Min = 29 – 4 = 25) D. Median (I would enter data into Excel to 4 0 4679 find the Median is 15 or do by ordering (3) 1 557 2 2 49 data and identifying) E. The range of the data that would contain 68% of the results. (Mean – Std Dev, Mean + Std Dev) which is (14 – 8.47, 14 + 8.47) or (5.53,22.47)
  • 7. GM 533 Final Exam Prep Here I used =MEDIAN(A1:A9) It returned the median of 15.
  • 8. GM 533 Final Exam Prep • Sample Question on Hypothesis Testing – Pepito’s Pizza Works is putting pizzas out by delivery as fast as they can. Pepito’s claims they can deliver pizzas within their delivery area in less than 29 minutes. You are given the following data from a sample. Sample size: 120 Deliveries Population standard deviation: 1.4 Sample mean: 28.3 Formulate a hypothesis test to evaluate the claim.
  • 9. GM 533 Final Exam Prep • Sample Question on Hypothesis Testing – Pepito’s Pizza Work is putting pizzas out by delivery as fast as they can. Pepito’s claims they can deliver pizzas within their delivery area in less than 29 minutes. You are given the following data from a sample. Sample size: 120 Deliveries Population standard deviation: 1.4 Sample mean: 28.3 Formulate a hypothesis test to evaluate the claim. – Answer: Ho: µ ≥ 29, Ha : µ < 29 – (In this case, the claim was Ha) – Remember Ho always contains equality (It will either be =, ≤ or ≥) – Ha will be either ≠, < or >
  • 10. GM 533 Final Exam Prep • Confidence interval Example – Acme computers needs to find a new vendor for their hard drives. They are considering using Howie’s Hard Drives as a vendor. Acme’s requirement is that 95% of the hard drives last 24000 hours ± 2000 hours. The following data is from an independent source who evaluated Howie’s. Should Acme buy from Howie’s? Explain your answer. (Follows on next page)
  • 11. GM 533 Final Exam Prep • Mean = 24500 • Sample Standard Deviation 2250 • Min 21402 • Max 29463 • Margin of Error 4500 • Answer Follows
  • 12. GM 533 Final Exam Prep • No, Acme shouldn’t buy from Howie’s looking at their requirements (24000 – 2000, 24000 + 2000) which is (22000, 26,000). Based on the results given, Howie’s would yield a tolerance of (24500 – 2*2250, 24500+2*2250) which is (20000, 29000). This does not meet Acme’s requirement. You could also see this by looking at the margin of error.
  • 13. GM 533 Final Exam Prep • Example on Pivot/Contingency Tables • The table below gives the number of cars of various colors and the state tag on the car for a parking lot in a mall close to DC. VA MD DC Other State Total Blue 4 8 9 3 24 Black 9 7 11 8 35 White 12 14 21 15 62 Other 37 10 29 35 111 Total 62 39 70 61 232
  • 14. GM 533 Final Exam Prep • Based on the table, find the probability that a car is from VA or MD. • Based on the table, given that a car is from DC, find the probability it is black.
  • 15. GM 533 Final Exam Prep VA MD DC Other State Total Find the probability that a car is Blue 4 8 9 3 24 from VA or MD. Add 62 + 39 to get Black 9 7 11 8 35 So the answer would be 91/232 or it’s White 12 14 21 15 62 decimal form. Other 37 10 29 35 111 Total 62 39 70 61 232 62 + 39 = 91
  • 16. GM 533 Final Exam Prep VA MD DC Other State Total Given that a car is from DC, find Blue 4 8 9 3 24 the probability it is black. Black 9 7 11 8 35 Given that it is from DC means we are only White 12 14 21 15 62 dealing with the 70 cars from DC. Other 37 10 29 35 111 There are 11 of those that are black, so the Total 62 39 70 61 232 probability is 11/70
  • 17. GM 533 Final Exam Prep • Normal Distribution Example – The number of students who use the dining hall at an urban college on a given day is normally distributed with a mean of 1578 students and a standard deviation of 274 students.
  • 18. GM 533 Final Exam Prep • I’m suggesting Excel to work these types of problems even if you are given partial Minitab results. • Go to http://highered.mcgraw- hill.com/sites/0070620164/student_view0/exc el_templates.html And download the template titled Normal Distribution.
  • 19. GM 533 Final Exam Prep • It will look something like this when you open it
  • 20. GM 533 Final Exam Prep • Before doing anything else, click the “Review” tab at the top of Excel (Between Data and View), then click “Unprotect Sheet”. After clicking “Unprotect Sheet” it will say “Protect Sheet.” Leave it that way and save to your computer. You now have a cool Normal Distribution Calculator.
  • 21. GM 533 Final Exam Prep • Back to our problem…. • Questions • What is the probability that less than 1400 students will use the dining hall? • What is the probability that more than 1700 will use the dining hall? • What is the probability that between 1400 and 1600 students will use the dining hall? • Get that Normal Distribution Excel Calculator ready and be amazed!
  • 22. GM 533 Final Exam Prep
  • 23. GM 533 Final Exam Prep
  • 24. GM 533 Final Exam Prep This gives This gives you area to you area to the right the left based on based on your mean your mean and and standard standard deviation. deviation. This gives you area between two values based on your mean and standard deviation.
  • 25. GM 533 Final Exam Prep I entered I entered 1700 in the 1400 in the green cell green cell which gives which gives me the me the probability probability of more of less than than 1700 1400 students students using the using the dining hall. dining hall. The answer The answer is 0.3281 is 0.2580 I entered 1400 in the left cell and 1600 in the right green cell which gives me the probability of between 1400 and 1600 students using the dining hall. The answer is 0.2740
  • 26. GM 533 Final Exam Prep • Another Confidence Interval Example – I randomly sampled 18 engineers where I work and asked them how many projects they have worked on in the last five years. The sample mean was 21, with a standard deviation of 5. What is the mean number of projects of all engineers at my research center? Why? What is the 95% confidence interval for the population mean? You are given the information below from Minitab. One-Sample T N Mean StDev SE Mean 95% CI 18 21.00 5.00 1.18 (18.51, 23.49)
  • 27. GM 533 Final Exam Prep • Another Confidence Interval Example – I randomly sampled 18 engineers where I work and asked them how many projects they have worked on in the last five years. The sample mean was 21, with a standard deviation of 5. What is the mean number of projects of all engineers at my research center? Why? What is the 95% confidence interval for the population mean? You are given the information below from Minitab. Answer: 21 projects would be the best estimate for the mean. I would expect 95% of the population mean to fall between 18.51 and 23.49 projects. The t is used because of the sample size.
  • 28. GM 533 Final Exam Prep • Regression Example – I did an analysis to determine if the number of hours studied for a final exam related to the Final Exam grade for students. On the sheets that follow you will see what my Minitab results were.
  • 29. General Regression Analysis: Final Grade versus Hours of Analysis of Variance Study Source DF Seq SS Adj SS Adj MS F P Regression Equation Regression 1 8090.71 8090.71 8090.71 171.274 0.000000 Final Grade = 34.2845 + 1.45508 Hours of Study Hours of Study 1 8090.71 8090.71 8090.71 171.274 0.000000 Error 22 1039.25 1039.25 47.24 Coefficients Lack-of-Fit 18 936.58 936.58 52.03 2.027 0.259215 Pure Error 4 102.67 102.67 25.67 Term Coef SE Coef T P Total 23 9129.96 Constant 34.2845 3.38091 10.1406 0.000 Hours of Study 1.4551 0.11118 13.0872 0.000 Fits and Diagnostics for Unusual Observations Summary of Model Final Obs Grade Fit SE Fit Residual St Resid S = 6.87303 R-Sq = 88.62% R-Sq(adj) = 88.10% 24 19 37.1947 3.17993 -18.1947 -2.98609 R PRESS = 1426.21 R-Sq(pred) = 84.38% R denotes an observation with a large standardized residual.
  • 30. GM 533 Final Exam Prep Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 48.8353 2.41382 (43.8294, 53.8413) (33.7280, 63.9426) Values of Predictors for New Observations Hours of New Obs Study 1 10
  • 31. GM 533 Final Exam Prep I did an analysis to determine if the number of hours studied for a final exam related to the Final Exam grade for students. On the sheets that follow you will see what my Minitab results were. Answer the following questions. Determine the regression equation. What conclusions are possible using the meaning of bo (intercept) and b1 (regression coefficient) in this problem? What does the coefficient of determination (r-squared) mean? Calculate the coefficient of correlation and explain what it means. Does this data provide significant evidence (a=0.05) that the final exam grade is associated with the hours studied? Find the p-value and interpret. Determine the predicted grade for someone who spends 10 hours studying for the final exam. What is the 95% confidence interval for the score for spending 10 hours studying on the test? What conclusion is possible using this interval?
  • 32. GM 533 Final Exam Prep I did an analysis to determine if the number of hours studied for a final exam related to the Final Exam grade for students. On the sheets that follow you will see what my Minitab results were. Answer the following questions. Determine the regression equation. y= 34.2845 + 1.45508x What conclusions are possible using the meaning of bo (intercept) and b1 (regression coefficient) in this problem? For each hour of study the final grade is increased by about 1.5 points (1.45508). bo represents the y intercept or 34.2845 in our case. It is the score that a student could expect to get without studying. What does the coefficient of determination (r-squared) mean? The .886 means that 88.6 percent of the variability of the final grade can be explained by the number of study hours. The other 11.4% would be due to something else or be unexplained.
  • 33. GM 533 Final Exam Prep Calculate the coefficient of correlation and explain what it means. Square Root of (0.886) is 0.942 which is r, the correlation coefficient. With a value this close to one, we could say there is strong positive correlation. Does this data provide significant evidence (a=0.05) that the final exam grade is associated with the hours studied? Find the p-value and interpret. Yes, the p value was 0. If it were above 0.05, I would have said “no.” Determine the predicted grade for someone who spends 10 hours studying for the final exam. 48.8353 What is the 95% confidence interval for the score for spending 10 hours studying on the test? What conclusion is possible using this interval? (43.8294, 53.8413) We would be 95% confident that if someone studied for 10 hours they would score on average between those two values.
  • 34. GM 533 Final Exam Prep • Multiple Regression – Be able to identify the multiple regression output from a Minitab analysis • Equation • F value • p value • Confidence intervals • (Basically reading from Minitab output)
  • 35. GM 533 Final Exam Prep • I will post these charts in the “Stat Cave” at – www.facebook.com/statcave – Good Luck!