Unit 9 Final Project MM207 essay

Unit9 Final Project MM207

Unit9 Final Project MM207

Question1.

1a.Therelationship between G.P.A and SAT score is stronger than therelationship between Age and GPA. This can clearly be shown becausethe correlation coefficient of GPA and SAT score of 0.927 is greaterthan the correlation coefficient of Age and GPA of 0.131.This is theresult I expected because the age of a person does not really affecthis/her Gpa instead the SAT score does.

1b

Ther-value of Project score and the final exam score is 0.888.Thisindicates a strong positive correlation between the final exam scoreand the project score.

1c.Prediction of the project scores when the final exam score is 82

Theequation relating project score and final exam score is

Y=0.7762X+ 18.365

Projectscore=0 .7762 (Final exam score)+18.365

Projectscore=0.7762 (82) +18.365

Projectscore=82.01

Question2

Question2a

2b.

Thetwo numerical measures that offer the best information for comparingthe performance between the two assignments are the mean and thestandard deviation.The mean which is the average score ,in theproject score is greater than the mean score of final exam score,which is 80.53. This shows that students performed better in aproject than the final exam.The standard deviation of the projectscore of 13.33 was lower than that of the final score of 15.25. Thisindicates that the project scores are closer to their higher meanthan the final exam scores. Therefore, we conclude that the studentsperformance was higher in the project than in the final exam

2c.

Itemerges that the two measures numerically that provide the bestinformation on variation of data are standard deviation and range.Therange which is the difference between the maximum score and a minimumscore of Final exam which is 54 is greater than the range in theproject score wchich is 48.17. The standard deviation which is theamount of variation of each score from the mean score of final scoreof 15.25 is greater than the standard deviation of the project score.

Thisshows that there is a greater variation in the final exam score thanthere is in project score

2D.

Thefemale students did better than the male students in the final exams.This can be shown by the mean where the average score of females of82.588 was higher than the average score of males of 78.471.Themedian score of females of 87 is also higher than that of males of82.The standard deviation of females of 14.116 is lower than that ofmales of 16.47 which means that the female score didn’t vary muchfrom their higher mean.

Question3

3(a)

Usingexcel the linear equation relating age and the final exam score wasy=-0.70008X +99.079 where y was the final score and x was theage.using this the final score of Ron was 82 and that of Sally was79.

Calculatingthe z scores

Themean score for males was 78.47(S.D=16.47) and for females was82.588(S.D=14.116)

Zscore for roy =(82-78.47)/16.47=0.21

Zscore for sally=(79-82.588)/14.116=-0.25

Fromthe z scores roy was 0.21 standard deviations above the male studentsmean and sally was 0.25 standard deviations below the mean of femalestudents.This shows that sally is farther away from the female meanthan roy is from the male mean.

3B.There were a total of 34 students and 18 of them scored higher thanthe gpa score of sally of 3.35

The% of students who scored higher than saly’s gpa is(18/34)*100=52.94%..this shows that 52.94 students scored a highergpa than sally’s gpa of 3.35.

3C.

Therewas only one male student who scored a lower sat score than Ron’sSat score.The % male students who scored less than Ron’s SAT scoreis (1/17)*100=5.88%.

Ron’sscore is 800. Mean male SAT score is 1135.15 with a standarddeviation of 227.64.

Zscore=(800-1135)/227.64=-1.472.

Thismeans that Ron’s SAT score is 1.472 standard deviations below theaverage male SAT score. Using the Z table, we conclude that 92.92% ofthe male students performed better than Ron in the SAT test.

Question4

4(A)

Relativeand cumulative frequency chart of the 5 categories of GPA grades

4B)bar chart showing the frequencies of each letter grade

Question5

5A)

95%confidence interval for the true population mean time that allstudents spend studying per week

Samplemean=9 .469

Standarderror=0.679

Zscore =1.96

95%C.I=9.469+-1.96*0.679

=8.139and 10.799

95%confidence interval = [8.139 10.799]

A1.

Ifa student spends 15 hours per week studying, it is significantlydifferent from the population mean because it is outside the 95%confidence interval(Box, Hunter &amp Hunter,1978).

5B).

Theuse of sample statistics in estimation of the population parametersgiven that it is not feasible to measure the entire population. Anexample of this is the sample mean. We can do the example of thesample mean of the SAT score and estimate the population mean and its95% Confidence interval.

TheSAT score sample mean is 1159.617.This is the point estimate meanthat shows us that the average population SAT score is likely to be1159.617. The 95 % confidence interval (1159.617+-81.642) which is[1077.975 1241.259]. This shows us that we can be 95% confident thatthe average SAT score of the population is between 1077.975 and1241.249.

References

Box,G. E., Hunter, W. G., &amp Hunter, J. S. (1978). Statistics forexperimenters.