Project Part 1 –worth 10 of the 20% Project grade - DUE SUNDAY, April 2 In this part of the project, you will be calculating measures on your data set using both the raw-data and grouped-data formulas in Chapter 2. For the set of data sent to you, (a) find the mean, median and range of the data, treating it as raw data. (b) make a grouped frequency distribution table consisting of 6 - 8 classes (recall that the instructions are found on p. 38) (c) find the mean, modal class, variance and standard deviation from the grouped table - include all relevant columns of information you need ( x , xf , x  x, ( x  x ) 2 , ( x  x ) 2 f . Show work (if you make an error and I can’t see where you went wrong, then I can’t find your error!). After doing this, compare the mean you found from the table to the raw-data mean. Remember that since you use class midpoints as approximate data values, the grouped mean probably won’t equal the raw one, but I want to see how close the two measures are (in other words, how good the approximation is). (d) write THREE interpretations of the data from the table. In other words, what do you learn about the set of data in grouped (organized) form that wasn't apparent from the raw (unorganized) form? I DON”T want you to write general comments that describe the benefits of organizing data; rather, I want you to tell me what you learn about your particular set of data after having organized it. For example, if your set of data was a list of people’s heights, don’t tell me (for example) that organizing the data in a table helps you to see the distribution of heights; rather, tell me something (for example) like,”From the table I learned that the majority of people in the list had a height over 58 inches.”, a discovery that is not as obvious from an unorganized list of numbers! Save your work (please use a filename that begins with your last name and that indicates what the file is (ex. DePriter PART 1.doc) and email it to me (there is NO drop box in ulearn.). I also accept pdfs and scans (jpgs). Project Part 2 – worth 4 of the 20% Project grade - DUE SUNDAY, April 30 In this part of the Project, you will be converting your frequency distribution from Part 1 into a discrete probability distribution. Using your class midpoints as X values, determine the "relative frequency" of each class by dividing its frequency by n. These will serve as the P(x)'s. Make a probability distribution table with all necessary columns ( xP ( x ), x   , ( x deviation. SHOW WORK.   ) 2 , P ( x )( x   ) 2 ) and calculate the mean, variance and standard How do the values of these measures compare to those you calculated in Part 1? Project Part 3 – worth 6 of the 20% Project grade - DUE SATURDAY, May 18 In the final part of the project, you will conduct two hypothesis tests for the population mean based on your data. BEYOND THE BASICS… Chapter 2, Part 3 Covariance and Correlation College of Arts & Sciences Department of Mathematics MATH5100: Statistical Methods Covariance and Correlation Covariance and correlation are measures used to determine the association between two variables, and how strong that association is. Given two samples of data represented by x and y, covariance   x  x y  y   cov( x, y)  n 1 cov( x, y) correlatio n coefficien t  corr( x, y)  sx s y where sx and sy are the standard deviations of the x and y values, respectively. 2 Covariance and Correlation A stock broker studies closing returns for two stocks over a 5day period. The data is summarized in the following table: Day Stock X (%) Stock Y (%) 1 0.9 1.6 2 2.1 2.4 3 3.2 2.7 4 1.6 2.0 5 2.2 2.3 3 Covariance and Correlation A stock broker studies closing returns for two stocks over a 5day period. The data is summarized in the following table: Day Stock X (%) Stock Y (%) 1 0.9 1.6 2 2.1 2.4 3 3.2 2.7 4 1.6 2.0 5 2.2 2.3 covariance  cov( x, y )   x  x y  y  The statistics for these samples are X : x  2.0, s  0.85 Y : x  2.2, s  0.42 n 1 (0.9  2.0)(1.6  2.2)  (2.1  2.0)( 2.4  2.2)  (3.2  2.0)( 2.7  2.2)  (1.6  2.0)( 2.0  2.2)  (2.2  2.0)( 2.3  2.2)  5 1 (1.1)( 0.6)  (0.1)(0.2)  (1.2)(0.5)  (0.4)( 0.2)  (0.2)(0.1)  4 0.66  0.02  0.60  0.08  0.02 1.38    0.345 4 4 4 Covariance and Correlation A stock broker studies closing returns for two stocks over a 5day period. The data is summarized in the following table: Day Stock X (%) Stock Y (%) 1 0.9 1.6 2 2.1 2.4 3 3.2 2.7 4 1.6 2.0 5 2.2 2.3 covariance  cov( x, y )  A positive covariance of 0.345 tells us that the two stocks “moves” in the same way: if the return for one increases, then the other’s return probably will as well.  x  x y  y  n 1 (0.9  2.0)(1.6  2.2)  (2.1  2.0)( 2.4  2.2)  (3.2  2.0)( 2.7  2.2)  (1.6  2.0)( 2.0  2.2)  (2.2  2.0)( 2.3  2.2)  5 1 (1.1)( 0.6)  (0.1)(0.2)  (1.2)(0.5)  (0.4)( 0.2)  (0.2)(0.1)  4 0.66  0.02  0.60  0.08  0.02 1.38    0.345 4 4 5 Covariance and Correlation A stock broker studies closing returns for two stocks over a 5day period. The data is summarized in the following table: Day Stock X (%) Stock Y (%) 1 0.9 1.6 2 2.1 2.4 3 3.2 2.7 4 1.6 2.0 5 2.2 2.3 The statistics for these samples are X : x  2.0, s  0.85 Y : x  2.2, s  0.42 correlatio n coefficien t cov( x, y )  corr( x, y )  sx s y 0.345 0.345    0.97 (0.85)(0.42) 0.357 6 Covariance and Correlation A stock broker studies closing returns for two stocks over a 5day period. The data is summarized in the following table: Day Stock X (%) Stock Y (%) The correlation coefficient, which 2 2.1 2.4 always equals a value between -1 3 3.2 2.7 and 1, tell us that the relationship 4 1.6 2.0 between Stocks X and Y is very 5 2.2 2.3 strong and “positive”: as one increases so will the other. A value correlatio n coefficien t close to 1 indicates this; a negative cov( x, value y ) indicates that they move in  corr( x, y )  s x s y opposite directions; a value close to 0 indicates random, non0.345 0.345 corresponding movement.    0.97 1 0.9 (0.85)(0.42) 1.6 0.357 7 Covariance and Correlation If the X-Y pairs are sketched on a coordinate system, the possible relationships are sketched below for various values of the correlation coefficient. 8 BEYOND THE BASICS… Chapter 2, Part 2 Other Types of Means College of Arts & Sciences Department of Mathematics MATH5100: Statistical Methods Two other types of means that can be calculated on sets of quantitative data are the geometric mean and the harmonic mean. The geometric mean of two numbers, a and b, is the square root of ab, geometric mean  ab The geometric mean of three numbers, a, b, and c, is the cube root of abc, geometric mean  3 abc , and so on. 2 Two other types of means that can be calculated on sets of quantitative data are the geometric mean and the harmonic mean. The geometric mean of two numbers, a and b, is the square root of ab, geometric mean  ab The geometric mean of three numbers, a, b, and c, is the cube root of abc, geometric mean  3 abc , and so on. Geometric means are useful when you want to compare sets of data with more than one quantitative attribute that you want to consider. 3 For example, suppose you want to see an action movie this weekend and there’s a choice of two: one that had an opening attendance of 28,000,000 and an audience rating of 7.2 out of 10, and another with an opening attendance of 19,000,000 and a rating of 9.1. Taking this information into account, which movie appears to be the better choice to see? Despite the large differences in attendance and ratings, the geometric mean will provide a good comparison. 4 For example, suppose you want to see an action movie this weekend and there’s a choice of two: one that had an opening attendance of 28,000,000 and an audience rating of 7.2 out of 10, and another with an opening attendance of 19,000,000 and a rating of 9.1. Taking this information into account, which movie appears to be the better choice to see? The geometric means for the two movies are (28,000,000)(7.2)  201,600,000  14,198.59 (19,000,000)(9.1)  172,900,000  13,149.14 Taking into consideration both attendance and rating, the first movie with the larger geometric mean appears to be the better choice. 5 Geometric means are also better to find when quantitative values are not independent. This is the case with percentage figures over successive periods of time like annual investment returns. One year’s return affects the next because it determines how much can be invested the next year. Suppose returns for a certain investor over the past four years were 50%, 10%, -70% and 40%. Determine the average return. 6 Geometric means are also better to find when quantitative values are not independent. This is the case with percentage figures over successive periods of time like annual investment returns. One year’s return affects the next because it determines how much can be invested the next year. Suppose returns for a certain investor over the past four years were 50%, 10%, -70% and 40%. Determine the average return. We need to add 1 to each return in order to avoid having a negative value within the root. We compensate for this by subtracting 1 from the root afterward. The geometric mean is 4 (1.50)(1.10)(0.30)(1.40)  1  4 0.693  1  0.912  1  0.088 7 Geometric means are also better to find when quantitative values are not independent. This is the case with percentage figures over successive periods of time like annual investment returns. One year’s return affects the next because it determines how much can be invested the next year. We need to add 1 to each return in order to avoid having a negative value within the root. We compensate for this by subtracting 1 from the root afterward. The geometric mean is 4 (1.50)(1.10)(0.30)(1.40)  1  4 0.693  1  0.912  1  0.088 Note that if we added these returns and divided by four, the mean equals +7.5%! In reality, the rate of -70% has a much more profound effect on the “average”. 8 The harmonic mean of n quantities, x 1 , x 2 , x 3, … x n , is harmonic mean  n 1 1 1 1    ... x1 x2 x3 xn It is used to find the average of rates or ratios. For example, if you drive a certain distance at a velocity of 60 mph, and then drive the same distance at a velocity of 40 mph, then your average velocity is 48 mph. 2 2 2 2(120)     48 1 1 2 3 5 5   60 40 120 120 120 9 The harmonic mean of n quantities, x 1 , x 2 , x 3, … x n , is harmonic mean  n 1 1 1 1    ... x1 x2 x3 xn It is used to find the average of rates or ratios. 2 1 1  60 40  2 2 3  120 120  2 2(120)   48 5 5 120 Why is this correct? Suppose you drove for 240 miles at each velocity. Then you drove for 4 hours at 60 mph and for 6 hours at 40 mph. Therefore, you drove 480 miles in 10 hours, at an average velocity of 48 mph. Note that simply averaging 60 and 40 and getting 50 mph is not correct. 10 The harmonic mean of n quantities, x 1 , x 2 , x 3, … x n , is harmonic mean  n 1 1 1 1    ... x1 x2 x3 xn It is used to find the average of rates or ratios. 2 1 1  60 40  2 2 3  120 120  2 2(120)   48 5 5 120 Why is this correct? Suppose you drove for 240 miles at each velocity. Then you drove for 4 hours at 60 mph and for 6 hours at 40 mph. Therefore, you drove 480 miles in 10 hours, at an average velocity of 48 mph. Note that simply averaging 60 and 40 and getting 50 mph is not correct. 11 The harmonic mean of n quantities, x 1 , x 2 , x 3, … x n , is harmonic mean  n 1 1 1 1    ... x1 x2 x3 xn It is used to find the average of rates or ratios. In finance, the harmonic mean is used, for example, to find averages of interest rates and price-earnings ratios, 12 BEYOND THE BASICS… Chapter 2, Part 1 Another Type of Ogive College of Arts & Sciences Department of Mathematics MATH5100: Statistical Methods Another column of information that can be calculated from a grouped frequency distribution is cumulative relative frequency. For each class, cumulative frequency cumulative relative frequency  sample size 2 cumulative frequency cumulative relative frequency  sample size In the following distribution, this formula has been used to fill the last column. Class Boundaries Frequency Relative Frequency Cumulative Frequency Cumulative Relative Frequency 0.5 – 9.5 16 16/60 = 0.267 16 16/60 = 0.267 9.5 – 18.5 23 23/60 = 0.383 16 + 23 = 39 39/60 = 0.65 18.5 – 27.5 11 11/60 = 0.183 39 + 11 = 50 50/60 = 0.833 27.5 – 36.5 5 5/60 = 0.083 50 + 5 = 55 55/60 = 0.917 36.5 – 45.5 3 3/60 = 0.050 55 + 3 = 58 58/60 = 0.967 45.5 – 54.5 2 2/60 = 0.033 58 + 2 = 60 60/60 = 1.000 n = 60 3 One use of this information is to produce a cumulative relative frequency ogive, depicted below from this distribution. Cumulative Relative Frequency Class Boundaries 4 One use of this information is to produce a cumulative relative frequency ogive, depicted below from this distribution. Cumulative Relative Frequency Class Boundaries This ogive has the same shape as one that was made with cumulative frequencies, but interpretations using percentages can be drawn from this version. 5 One use of this information is to produce a cumulative relative frequency ogive, depicted below from this distribution. Cumulative Relative Frequency .46 15 25 Class Boundaries 75% of the data is approximately 25 or less. Approximately 46% of the data is 15 or less. 6 BEYOND THE BASICS… Chapter 2, Part 2 Medians of Distributions College of Arts & Sciences Department of Mathematics MATH5100: Statistical Methods Median of a data set (ungrouped frequency distribution) The median of an ungrouped frequency distribution may be difficult to find if the sample or population size is large. If so, the following aid can be used to simplify the process: The median is the value in the n 1 N 1 th or th position of the ordered list. 2 2 2 Median of a data set (ungrouped frequency distribution) The median is the value in the n 1 N  1 position of the ordered list. or th th 2 2 In this sample, the median is the x f 3 16 4 24 5 11 6 26 n 1 77  1 th  th  39th 2 2 value Finding the cumulative frequencies can help identify the median: they are, in order, 16, 40, 51 and 77, so the 39th value is a 4. n = 77 3 Median of a data set (ungrouped frequency distribution) The median is the value in the n 1 N  1 position of the ordered list. or th th 2 2 In this population, the median is the x f 213 2016 214 524 215 3481 216 127 N = 6148 N 1 6148  1 th  th  3074.5th 2 2 value Finding the cumulative frequencies can help identify the median: they are, in order, 2016, 2540, 6021 and 6148, so the 3074.5th value is a 215 (halfway between the 3074th and 3075th). 4 Median of a data set (grouped frequency distribution) The median of a grouped frequency distribution is fairly complicated to find. The formula is i 50% of n  cf b  median  L  fm (note that for a population, N replaces n). L is the left boundary of the class containing the median i is the class width fm is the frequency of the class containing the median cfb is the cumulative frequency of the class before the class containing the median 5 Median of a data set (grouped frequency distribution) i 50% of n  cf b  median  L  fm Class Boundaries Frequency Relative Frequency Cumulative Frequency Cumulative Relative Frequency 0.5 – 9.5 16 16/60 = 0.267 16 16/60 = 0.267 9.5 – 18.5 23 23/60 = 0.383 16 + 23 = 39 39/60 = 0.65 18.5 – 27.5 11 11/60 = 0.183 39 + 11 = 50 50/60 = 0.833 27.5 – 36.5 5 5/60 = 0.083 50 + 5 = 55 55/60 = 0.917 36.5 – 45.5 3 3/60 = 0.050 55 + 3 = 58 58/60 = 0.967 45.5 – 54.5 2 2/60 = 0.033 58 + 2 = 60 60/60 = 1.000 median in here Let us assume this distribution represents a sample. The first thing we need to determine is which class contains the median. Look for 0.500 in the cumulative relative frequency column (this would represent the halfway point into the distribution). IF IT IS NOT THERE, the median is in the class with the NEXT HIGHEST CUMUL. REL. FREQ. 6 Median of a data set (grouped frequency distribution) i 50% of n  cf b  median  L  fm Class Boundaries Frequency Relative Frequency 0.5 – 9.5 16 16/60 = 0.267 23 18.5 – 27.5 i 9( cfb Cumulative Relative Frequency 16 16/60 = 0.267 23/60 = 0.383 16 + 23 = 39 39/60 = 0.65 11 11/60 = 0.183 39 + 11 = 50 50/60 = 0.833 27.5 – 36.5 5 5/60 = 0.083 50 + 5 = 55 55/60 = 0.917 36.5 – 45.5 3 3/60 = 0.050 55 + 3 = 58 58/60 = 0.967 45.5 – 54.5 2 2/60 = 0.033 58 + 2 = 60 60/60 = 1.000 9.5 – 18.5 L Cumulative Frequency fm median in here n = 60 L is the left boundary of the class containing the median i is the class width fm is the frequency of the class containing the median cfb is the cumulative frequency of the class before the class containing the median 7 Median of a data set (grouped frequency distribution) i 50% of n  cf b  median  L  fm Class Boundaries Frequency Relative Frequency 0.5 – 9.5 16 16/60 = 0.267 23 18.5 – 27.5 i 9( cfb Cumulative Relative Frequency 16 16/60 = 0.267 23/60 = 0.383 16 + 23 = 39 39/60 = 0.65 11 11/60 = 0.183 39 + 11 = 50 50/60 = 0.833 27.5 – 36.5 5 5/60 = 0.083 50 + 5 = 55 55/60 = 0.917 36.5 – 45.5 3 3/60 = 0.050 55 + 3 = 58 58/60 = 0.967 45.5 – 54.5 2 2/60 = 0.033 58 + 2 = 60 60/60 = 1.000 9.5 – 18.5 L Cumulative Frequency fm median in here n = 60 9 median  9.5  50% of 60  16 23 9 9  9.5  (30  16)  9.5  (14)  9.5  5.48  14.98 23 23 8 Median of a data set (grouped frequency distribution) Suppose 0.500 is IN the cumulative relative frequency column? The median is the right boundary of the class with the 0.500. This is illustrated in the example on the next slide. 9 Median of a data set (grouped frequency distribution) i 50% of n  cf b  median  L  fm L i 9( Class Boundaries Frequency Relative Frequency 0.5 – 9.5 16 16/60 = 0.267 14 18.5 – 27.5 Cumulative Frequency cfb Cumulative Relative Frequency 16 16/60 = 0.267 14/60 = 0.233 16 + 14 = 30 30/60 = 0.500 11 11/60 = 0.183 30 + 11 = 41 41/60 = 0.683 27.5 – 36.5 5 5/60 = 0.083 41 + 5 = 46 46/60 = 0.767 36.5 – 45.5 13 13/60 = 0.217 46 + 13 = 59 59/60 = 0.983 45.5 – 54.5 1 1/30 = 0.033 59 + 1 = 60 60/60 = 1.000 9.5 – 18.5 fm n = 60 median in here Right boundary of the class with the 0.500 ✓ 9 median  9.5  50% of 60  16 14 9 9  9.5  (30  16)  9.5  (14)  9.5  9  18.5 14 14 10
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