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Midterm

# QMS102 Midterm: QMS 102 - Midterm Notes

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School
Ryerson University
Department
Quantitative Methods
Course
QMS 102
Professor
Douglas Mc Kessock
Semester
Fall

Description
Which is a better measure of Central Tendency: mean or median? 10% rule: 1) Calculate the difference A=|Mean-Median|. 2) Calculate the ―10% of the smaller of the mean and median‖, say B. 3) If AB, then median is the preferred measure. ~ mean x 11522.4615 , and median x  4667 . Now we apply the 10% rule: The difference A=| 11522.4615  4667 | 6855.4635 , and the 10% of the smaller, B=10%*4667=466.7. Since A>B, we conclude that the median is a better measure! Mode is data entry that occurs the most 1) Arrange data in an ascending data array Find the value . k when r=integer, k x r; when r=m+1/2, Pk (x  m m1) 2 Find the rank of the 39 percentile: n=13, k=39. So, r = HalfRound(nk/100+1/2) = HalfRound(5.57)=5.5 So, P  (x  x )  (2752  3646 )  3199 39 2 5 6 2 Interquartile Range = Q3 – Q1 Range : Highest number of data set – lowest number For ABC Inc., x  32.21875 and s x 6.735 . So, CV= s x x *100%  21% ; For XYZ Inc., x  32.1023 and sx 6.0278 . So, CV= sx/ x *100% 19% . So, we can conclude that ABC Inc. has more variability. We can use a CASIO calculator to compute the standard deviation:  x for the population standard deviation, and sx for the population standard deviation. (2) To find the variance, we need to square the standard deviation. You were asked to fill in a survey by Ryerson University about the number of courses you intend to take. One of the questions asked is as follow: How many courses do you intend to take in the Winter 2010? Refer to the question above being asked on the survey form. What is the measurement scale of the response data? (a) Nominal (b) Ordinal (c) Interval - Continuous (d) Ratio – Discrete (e) Ratio - Continuous LIF=Q1-1.5*IQR; RIF=Q3+1.5*IQR; LOF=Q1-3*IQR; ROF=Q3+3*IQR Note: (1) LIF=Left Inner Fence; LOF=Left Outer Fence; (2) RIF=Right Inner Fence; ROF=Right Outer Fence; (3) All data values that fall between the fences are call outliers, using ―o‖ to indicate them; All data values that lie beyond the outer fences are call extremes, using ―*‖ to indicate them. To measure the skewness, one can use the following: 3(mean-median) Skewness= ----------------------------- standard deviation 1) if meanmedian, then skewness>0 (positively skewed, skewed right); 3) if mean=median, then skewness=0 (symmetric). . The quartiles are the values that occupy the quarter positions: First quartile,Q 1 P 25 Second quartile, Q 2 P 50 Third quartile, Q 3 P 75 A five-number summary that consist of X smallest, Q 1 , Median, Q 3 , X largest Class Width = Highest Number - Lowest Number / 5 Right Whisker = greater than Q3 but less than RIF. Value close to RIF Left Whisker = greater than Q1 but less than LIF . Value Close to LIF How to calculate control limits? Assume that we collect k samples of size n. Lebe the range of the ith sample, i=1,2,…, k. Then i k R  k  R i (Rk..1  R ) k i1 UCL RD  R4 LCL RD  R3 A Normal distrubtion with mean of 0 and standard deviation of 1 Example 5. Approximate the mean of data: Height in inches. Class frequency Middle point 50—52 5 51 53—55 8 54 56—58 12 57 59—61 13 60 62—64 11 63 Solution: x  (5*518*54 12*57 13*60 11*63)/58 12 13 11)  58.041 Job classification # of employees Annual Salary (\$000’s) Senior Management 3 52 Middle Management 10 38 Cooking staff 45 20 Serving staff 74 18 Maintenance staff 7 23 What’s the mean annual salary? x  (3*5210*38 45*2074*187*23)/(310 45747) Solution:  21.1 Tom has done 10 weekly assignments through the whole semester, and received an average ―83‖. However, Tom realized that the Professor did not correct one of his grade for the assignment #8. It was supposed to be 95, but he originally received 80. Suppose this correction has been
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