1BUS 304 Homework Assignment #2, Due date: July 18, 2008 Submitted by: (sign) Submission date: (sign) Problem #1: Given the following data on the gasoline prices, construct a bar chart that displays the data effectively. Year Average Unleaded Price Average Premium Price 1999 $1.22 $1.34 2000 $1.29 $1.39 2001 $1.37 $1.48 2002 $1.21 $1.31 Problem #2: Construct a line chart for the above figure. (The data should show two-lines for Unleaded and Premium, respectively) (Both charts can be used. But you may note that the second one shows a better difference.)2Problem #3: (Textbook 2.23, P55) Problem #4: (Textbook 2.42 (b) and (c), P64) (By adding a trend line, (right-click the scattered dots, you see the option “adding a trend line”) the relationship presented much better.) X-axis: overtime hours worked (Independent variable, the reason to the variation of Y, see chapter 12) Y-axis: performance rating (Dependent variable, the result, see chapter 12) The more hours you’ve worked overtime, the higher rating will be. Positive correlation.3Problem #5: (Textbook 2.9) Each month the American Automobile Association (AAA) generates a report on gasoline prices that they distribute to the newspapers throughout the state. On February 17th, AAA called a random sample of 51 stations to determine the price of unleaded gasoline that day. The resulting data are shown as follows: 1.07 1.31 1.18 1.01 1.23 1.09 1.29 1.10 1.16 1.08 0.96 1.66 1.21 1.09 1.02 1.04 1.01 1.03 1.09 1.11 1.11 1.17 1.04 1.09 1.05 0.96 1.32 1.09 1.26 1.11 1.03 1.20 1.21 1.05 1.10 1.04 0.97 1.21 1.07 1.17 0.98 1.10 1.04 1.03 1.12 1.10 1.03 1.18 1.11 1.09 1.06 a) Construct two histograms for this data set, using 5 classes for the first and 15 classes for the second. Use 0.95 as the lower limit of the first class. i) Range of Prices (rounded): min: $0.95 max: $1.70 Æ 1.7-.95 = 0.75 ii) 5 classes: Range covered by each bar: 0.75/5 = 0.15 a. Ranges: “0.95~1.09”, “1.10~1.24”, “1.25~1.39”, “1.40~1.54”, “1.54~1.69” b. Count the number of prices that falls into the range. Create the frequency table: (you should always double check the frequencies) Price Range ($) Frequency 0.95~1.10 271.10~1.25 191.25~1.40 41.40~1.55 01.55~1.70 1>=1.70 0 c. Graph: iii) 15 classes: range covered by each bar: 0.75/5 = 0.05 a. Ranges: “0.95~0.99”, “1.00~1.04”, “1.05~1.09”, “1.10~1.14”, …, “1.65~1.69”, .4b. Count the number of prices that falls into the range. Create the frequency table: Price Range ($) Frequency 0.95~1.00 41.00~1.05 111.05~1.10 121.10~1.15 91.15~1.20 51.20~1.25 51.25~1.30 21.30~1.35 21.35~1.40 01.40~1.45 01.45~1.50 01.50~1.55 01.55~1.60 01.60~1.65 01.65~1.70 1>=1.70 0 c. Graph: b) A local radio station has reported that 30% of the gas stations are charging $1.15 a gallon or more for gasoline. (1) Use one of the histograms you have produced to respond to this report. (2) Which histogram did you use? Why? Use the second graph, you can find the bars that are higher than $1.15. Summing up all the frequencies, you get 5+5+2+2+1=15. Percentage: 15/51=29.41%. So the radio station is very close. $1.15 or higher5Problem #6 Five wheat farms have been selected at random from those in a particular county. The following crop yields (total bushels of wheat) are given for each of the five, along with the number of acres on each farm. Yield 15030 43400 10260 13200 89200 Acres 80 60 75 55 140 a) Compute the mean yield per farm for this sample ()Total YieldMean Yield per FarmNumber of Farms15030 43400 10260 13200 89200534218 bushels of wheat=++++== b) Compute the mean yield per acre for this sample. ()Total YieldMean Yield per Acre=Total Acres15030 43400 10260 13200 8920080 60 75 55 140417.2927 bushels of wheat++++=++++= c) If the governor is considering the necessity of adopting new technologies which can improve the productivity, which mean yield does he care the most? Mean Yield per Acre measures productivity. It indicates how much on average each unit of land produces. Problem #7 (Book 3.26, page 103) (Hint, check whether it is sample or population data) Mean = 16.1 Std. Dev. = 3.54 (note, it is sample) No. of Visit > Mean + Std. Dev. = 19.64 Æ at least 20 times. Problem #8 (Book 3.12, page 93) question a) b) and d). Percentile Location Value 20th i= *2024(1)100+= 5 =19 (5th number) 25th i= *2524(1)100+= 6.25 = 20*(1-0.25)+ 23* 0.25 =20.75 (your final result) (location 6.25 indicates it is between 6th number and 7th number. Take the 6th number, 20, and the 7th number, 23, and use weight 75:25 to compute the weighted average gives 20.75. Note: the reason to use weights 75:25 is because the decimal of the location is 0.25, which is always the weight to the higher number.630th i= *3024(1)100+= 7.5 = 23*(1-0.5)+ 23* 0.5 =23 (your final result) (location 7.5 indicates to it is between 7th and 8th number. Decimal 0.5 indicates fifty-fifty weights. However, both 7th and 8th numbers are 23. Therefore, final result is 23.) 50th i= *5024(1)100+= 12.5 27.5 (average between 12th and 13th number. You should know why.) 75th i= *7524(1)100+= 18.7544.75 = 44*(1-0.75)+ 45* 0.75 90th i= *9024(1)100+= 22.5 85 = 70*(1-0.5)+ 100* 0.5 Problem #9 A golf club has 15 members in Aug 2005. Below is the aggregate information of the members’ annual income: Mean annual income: $80,000 Median annual income: $60,000 Mode annual income: $50,000 (3 members) Maximal annual income: $120,000 Minimal annual income: $40,000 a) Are the information above population parameters or sample statistics. Explain. It says the club only has 15 members. The above number is calculated from them. Therefore, those are Population Parameters. b) Is the data symmetric, right skewed or left skewed? Explain. Right-skewed. Because Mode<Median<Mean. c) In Sep 2005, a new member joined the club. His annual income is $200,000. What is the new mean annual income in this club now?(3 points) And what is the new range of income? (3 points) And the new mode? (3 points) After the joining of this new member, the club has 16 members. Since the new person’s income is $200,000, which is way higher than all the other people, so the maximal changes to $200,000 • Mode will not change because this person is obviously an “outlier”. The “highest frequency” income is still $50,000. • New Range = New Maximal – New minimal = $200K-$40K =$160K • New Mean = New total income /16 =Old