251y0213 10/07/02 Part I. ECO251 QBA1 FIRST HOUR EXAM OCTOBER 1, 2002 Name __________________ SECTION MWF 10 11 TR 11 12:30 (10 points) 1. (Lind et. al.) A student takes a survey of heights of 450 college women and divides them into 10 classes. Her first two class midpoints are 62.5" and 65.5". There are 91 women in the first class and 110 women in the second class. (5) a. What is the (width of the) class interval w ? b. What is the lower limit of the third class? c. What is the relative frequency of the second class? d. What is the cumulative relative frequency of the second class? e. If this distribution is skewed to the left, about what percent of the data is above the median? Solution: a) Since 65.5 - 62.5 = 3, the interval must be 3. b) The lower limit must be 67. If the class interval is 3, the class must extend by 1.5 on either side of the midpoint. The layout for the first three classes is given in the table below. n 450 f Class Midpoint F f Frel Fn f rel n 61 - 64 62.5 91 .2022 91 .2022 64 - 67 65.5 110 .2444 201 .4467 67 - 70 68.5 ? ? ? ? c) The relative frequency is .275 d) The cumulative relative frequency is .5025, which might be found as (about) the sum of .2022 and .2444. e) the median is defined as a point with 50% of the data above or below it. 2. Indicate whether the following are: Nominal Data, Ordinal Data, Interval Data, Continuous Ratio Data or Discrete Ratio Data. (3) a. A recent cartoon suggested that the numbers on pro Football jerseys be replaced by their salaries. What kind of data would the numbers be before the change? Ans: Nominal. b. What kind of data would the numbers usually be considered after the change? Ans: Continuous ratio. c. What kind of data is a team's score at half time? Ans: Discrete ratio. 3. All my family doctor's patient files are coded as follows: FS (Adult females who currently smoke); FN (Adult females who do not currently smoke); MS (Adult males who currently smoke) and MN (Adult males who do not currently smoke). Are these categories mutually exclusive and collectively exhaustive? If you say 'no' to either characteristic, explain. (2) Ans: The categories are mutually exclusive ( for instance no person who is coded FS will be in FN, MS or MN), but not collectively exhaustive because there is no reason to assume that all the family doctor's patients are adults. 1 251y0213 9/30/02 Part II. Compute an appropriate answer, showing your work (15 Points maximum - if you do more than 15 points, only your right answers will be counted.): 1) A sample of pipe outside diameters gives a mean of 16.0 inches and a standard deviation of 0.2 inches. a) If the median diameter is 15.9 inches and the mode is unknown (i) What is the maximum fraction of the pipes that could have a diameter above 16.4 x x 16 .4 16 2 . We know from the inches? (1) Ans: Get a z-score. k z s 0.2 Chebyshef rule that the fraction in the tail above k is less than 1 2 . Since k k 2, this fraction is below 1 4 25%. (ii) Between what two diameters must at least 15/16 of the pipe diameters lie? (1) Ans: If only 116 1 k 2 are outside the interval, we must have k 2 16 or k 4. Thus the interval is k 16.0 4.2 16.0 0.8 or 15.2 to 16.8. (iii) Is this distribution skewed to the left or the right, or is it symmetrical? (1) Ans: Since the median is below the mean, it should be skewed to the right. b) If, instead, the median diameter is 16.0 inches and the mode is also 16.0 inches, between what two diameters must almost all the pipe diameters lie? (1) Ans: Since 16 30.2 16 0.6 is 3 standard deviations from the mean, the Empirical Rule (which applies because the mean, median and mode are equal) says that there will be almost none outside 15.4 to 16.6. c) What is the coefficient of variation for this sample of pipes? (1) Ans: C s x 0.416 0.025 . 2) The newest computer in the headquarters of a firm that you are liquidating is five months old and the oldest is 99 months old. a) If the (absolute) frequencies are to be presented in a line graph in seven classes, what intervals would you use? Explain your reasoning using an appropriate formula and use it to fill in the table below.(3) 99 5 13 .43 so use 14. This is only a suggestion. Any number, like 15, somewhat above Ans: 7 13.57 will work, as long as you cover the range. Two possibilities are shown below. You should have shown only one. Class A B C D E F G From 3 17 31 45 59 73 87 to 16.9 30.9 44.9 58.9 72.9 86.9 100.9 From 0 15 30 45 60 75 90 to 14.9 29.9 44.9 59.9 74.9 89.9 104.9 b) What is the name of the type of graph that you are drawing (Is it a histogram?) and what would the x and y coordinates be of the last point on the line that you draw to represent the frequencies? (2) Ans: The graph is a frequency polygon and we must create an empty class to end it. For the first classification above, the interval is 14 and the midpoint of the last class on the table is 94, so the last point is x 108 , y 0 . For the second classification above, the interval is 15 and the midpoint of the last class on the table is 97.5, so the last point is x 112 .5, y 0 . 2 251y0213 9/30/02 3) For the numbers 70, 270, 470 and 670, compute the a) Geometric Mean b) Harmonic mean, c) Rootmean-square (2 each). Label each clearly. If you wish, d) Compute the geometric mean using natural or base 10 logarithms. (3 points if you need it here or two points if you need it in the next section - doing this is insurance, you cannot get more than cannot get more than 15 points on part II or 25 points on part III unless you do the extra credit in part III. ) x 1480 . This is not used in any of the following calculations and there is Solution: Note that no reason why you should have computed it! (a) The Geometric Mean. 1 x g x1 x 2 x3 x n n n 5951610000 x 4 70 270 470 670 4 5951610000 5951610000 0.25 277 .753 . Do any of you really believe that 5951610000 1 4 1 4 5951610000 ? 4 (b) The Harmonic Mean. 1 1 xh n 1 1 1 x 4 70 270 470 670 4 0.0142857 0.0037037 0.0021277 0.0014925 1 1 1 1 1 1 0.0216096 0.00540240 . So xh 1 185 .103 . 1 1 0.00540240 4 n x As I explained several times, 1 1 could not possibly be 1 1 . n 4 1480 x (c) The Root-Mean-Square. 1 1 1 2 x rms x 2 70 2 270 2 470 2 670 2 4900 72900 220900 448900 n 4 4 1 747600 186900 . 4 So x rms 1 n x 2 186900 432 .319 . (d) The geometric mean using logarithms. Using natural logarithms: 1 ln( x) 1 ln 70 ln 270 ln 470 ln 670 ln x g n 4 1 1 4.24850 5.59842 6.15273 6.50728 22 .5069 5.62673 . So 4 4 x g e 5.62673 277 .753 . Or using logarithms to the base 10: 1 log( x) 1 log70 log270 log470 log670 log x g n 4 1 1 1.84510 2.43136 2.67210 2.82607 9.77463 2.44366 . So 4 4 2.44366 x g 10 277 .753 . 3 251y0213 9/30/02 Part III. Do the following problems (25+ Points) 1. I have the following data for March electricity bills at a sample of 6 homes of similar sizes. 93 105 213 136 177 192 Compute the following: a) The Median (1) b) The Standard Deviation (4) c) The 3rd Quintile (2) Solution: Compute the Following: Note that x is in order Index 1 2 3 4 5 6 Total x 93 105 136 177 192 213 916 x2 8649 11025 18496 31329 36864 45369 151732 xx x x 2 -59.6667 3560.11 -47.6667 2272.11 -16.6667 277.78 24.3333 592.11 39.3333 1547.11 60.3333 3640.11 -0.0003 11889.33 Note that, to be reasonable, the mean, median and 3 rd decile must fall between 93 and 213. You should have done either the third column or the fourth and fifth columns. If you did both you were wasting time. n 6, x 916 , x 2 151732 , x x 0.00, x x 2 a) Just put the numbers in order and average the middle numbers, x.5 Or formally: position pn 1 a.b .57 3.5 11889.33 . x3 x 4 136 177 156 .5 . 2 2 x1 p xa .b( xa1 xa ) so x1.5 x.5 x3 0.5( x 4 x3 ) 136 0.5(177 136 ) 156 .5 . x 916 152 .667 b) x n x x 6 s 2 x 2 nx 2 n 1 151732 6152 .667 2 2377 .7 or 5 2 11889 .33 2377 .9 s 2377.9 48.76 n 1 5 c) The 3rd quintile has 60% below it. position pn 1 a.b 0.67 4.2 . a 4, .b 0.2 . s2 x1 p x a .b( x a1 x a ) so x1.6 x.4 x 4 0.2( x5 x 4 ) 177 0.2(192 177 ) 180 (New Formula: position 1 pn 1 a.b 1 0.6(5) 1 3.0 4.0 . a 4, .b 0.0 . x1 p xa .b( xa1 xa ) so x1.6 x.4 x 4 0.0( x5 x 4 ) 177 0.0(192 177 ) 177 ) 4 251y0213 9/30/02 2. A bus line takes a sample of the distances ridden by commuters and gets the following. is investigating the amount of time customers are put on hold when they call. The times are tabulated below. (Assume that the numbers are a sample.) a. Calculate the Cumulative Frequency (1) b. Calculate The Mean (1) amount frequency F c. Calculate the Median (2) 5 - 9.99 miles 5 5 d. Calculate the Mode (1) 10 - 14.99 miles 8 13 e. Calculate the Variance (3) 15 - 19.99 miles 9 22 f. Calculate the Standard Deviation (2) 20 - 24.99 miles 11 33 g. Calculate the Interquartile Range (3) 25 - 29.99 miles 16 49 h. Calculate a Statistic showing Skewness and 30 - 34.99 miles 19 68 Interpret it (3) 35 - 39.99 miles 12 80 i. Make an ogive of the Data (Neatness Counts!)(2) j. Extra credit: Put a (horizontal) box plot below the ogive using the same scale. Solution: x is the midpoint of the class. Our convention is to use x as the midpoint of 0 to 2, not 1.99999. If you did this using computational formulas, you should have the table below. 3 2 class x f Row fx fx fx 1 5-9.99 5 7.5 37.5 281.2 2109 2 10-14.99 8 12.5 100.0 1250.0 15625 3 15-19.99 9 17.5 157.5 2756.2 48234 4 20-24.99 11 22.5 247.5 5568.7 125297 5 25-29.99 16 27.5 440.0 12100.0 332750 6 30-34.99 19 32.5 617.5 20068.8 652234 7 30-39.99 12 37.5 450.0 16875.0 632812 Total 80 2050.0 58900.0 1809063 Definitional formulas give you the table below. There is no reason to do both the tables for computational and definitional formulas. Row 1 2 3 4 5 6 7 Total n class 5-9.99 10-14.99 15-19.99 20-24.99 25-29.99 30-34.99 35-39.99 f 80, fx f x x 2 x f 5 8 9 11 16 19 12 80 7.5 12.5 17.5 22.5 27.5 32.5 37.5 2050 , 6368.75, and fx 37.5 100.0 157.5 247.5 440.0 617.5 450.0 2050.0 2 58900 , f x x 3 f x x xx fx -18.125 -13.125 -8.125 -3.125 1.875 6.875 11.875 fx 3 -90.625 -105.000 -73.125 -34.375 30.000 130.625 42.500 0.000 1809063 , f x x 2 1642.58 1378.12 594.14 107.42 56.25 898.05 1692.19 6368.75 f x x 3 -29771.7 -18087.9 -4827.4 -335.7 105.5 6174.1 20094.7 -26648.4 f x x 0, 26648.4. Note that, to be reasonable, the mean, median and quartiles must fall between 5 and 40. a. Calculate the Cumulative Frequency (1): (See above - in red) The cumulative frequency is the whole F column. b. Calculate the Mean (1): x fx 2050 25.625 n 80 5 251y0213 9/30/02 c. Calculate the Median (2): position pn 1 .581 40.5 . This is above F 33 and below F 49, pN F .580 33 so the interval is 25-29.99. x1 p L p w so x1.5 x.5 25 5 27 .1875 16 f p d. Calculate the Mode (1) The mode is the midpoint of the largest group. Since 19 is the largest frequency, the modal group is 30 to 34.99 and the mode is 32.5. e. Calculate the Variance (3): s 2 s2 f x x n 1 2 fx 2 nx 2 n 1 58900 80 25 .625 2 6368 .75 80 .6171 or 79 79 6368 .75 80 .6171 79 f. Calculate the Standard Deviation (2): s 80.6171 8.97870 g. Calculate the Interquartile Range (3): First Quartile: position pn 1 .25 81 20.25 . This is above pN F F 13 and below F 22 so the interval is 15-19.99. x1 p L p w gives us f p .25 80 13 Q1 x1.25 x.75 15 5 15 3.8889 18 .8889 . 9 Third Quartile: position pn 1 .75 81 60.75 . This is above F 49 and below F 68, so the .75 80 49 interval is 30-34.99. x1.75 x.25 30 5 30 2.8947 27 .8947 . 19 IQR Q3 Q1 32.8947 18.8889 14.0058 . (New Formula: For the median - position 1 pn 1 1 0.579 40 .5 . This is the same result as on the previous page. For the first quartile - position 1 pn 1 1 0.2579 20.75 . This leads to the same result as above. For the third quartile - position 1 pn 1 1 0.7579 60 .25 . This leads to the same result as above.) h. Calculate a Statistic showing Skewness and interpret it (3): n k 3 fx 3 3x fx 2 2nx 3 80 1809063 325.625 58900 280 25.625 3 (n 1)( n 2) 7978 0.0129828 26648 .932 345 .97 . or k 3 or g 1 n (n 1)( n 2) k3 s 3 f x x 345 .97 8.9787 3 3 80 26648 .4 345 .97 79 78 0.4800 3mean mode 325 .625 32 .5 2.297 std .deviation 8.9787 Because of the negative sign, the measures imply skewness to the left. i. Make an ogive of the Data (Neatness Counts!)(2) An ogive is a line graph of the cumulative frequency. It should hit zero on the left at x 5 . The next point is 5 at x 10 . It continues to rise until x 40 , when the height is 80. It ends with a horizontal line. j. Your box plot should show the first and third quartiles at the left and right of the box and a vertical line across the box at the median. . It should be immediately below the ogive and use the same x points. or Pearson's Measure of Skewness SK 6