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1、Chapter1 &2 Class Name I .D. No. Marks Assignment 1 for Biostatistics (04110071)Due date: Mar.12, 20131. (2 points each) Judge the statement (true or false) and explain why if you think it is false.(1).In a pie chart, the area of a given sector is the number or proportional to the percentage of indi
2、viduals falling into that category.Judge: F Explain: In a pie chart, the area of a given sector is proportional to the percentage of individuals falling into that category.(2).The geometric mean is always less than the arithmetic mean if the data skewed to the right.Judge: T Explain:(3).The standard
3、 deviation may be regarded as an average of the deviations of the observations from the arithmetic mean.Judge: T Explain:(4).The mode often has a same value as the arithmetic mean and the median.Judge: F Explain: The mode often has a different value from both the arithmetic mean and the median.(5).H
4、ighly accurate data will always bring high level of precision.Judge: T Explain:(6).The arithmetic mean, geometric mean or variance all can be used as measurements of the central tendency of data.Judge: FExplain: Variance is used to measure the degree of data dispersion.(7).An extreme value of the ob
5、servation must be an outlier.Judge: FExplain: An extreme value sometimes is extreme results of variables inherent random variability, may be not an outlier.(8).Statistical diagrams, such as bar charts, histograms, and dynamic curve, can be used to show basic characters and changing patterns of any d
6、ata collected.Judge: FExplain: The bar chart is useful for categorical or discrete data; the histogram is useful for continuous data.(9).There must be a median existed in a data set.Judge: FExplain: If the number of data set observations is even, then the median lies midway between the central two o
7、bservations.(10).A systematic error is one in which the recorded value is consistently above (or below) its true value, so we can say human error is systematic error.Judge: FExplain: human error is random error.2. (2 points each) Please judge whether the following kinds of chart is an appropriate di
8、agram to show the frequency distribution of a discrete variable (true or false), and explain why if you think it is false. (1) .A histogram. Judge: F Explain: Histogram is used to show the frequency distribution of a continuous variable.(2) .A pie chart. Judge: T Explain: (3) .A bar chart. Judge: T
9、Explain: 3. (3 points each) Decide the type of following populations, in terms of real or hypothetical, dynamic or static, limited or unlimited, and explain why please. (1).The birth weights of all Chinese Holstein between 1990 and 2000 in Beijing. Type: Real; Static; Limited. Explain: The weights o
10、f all Chinese Holstein between 1990 and 2000 exist actually, so its real. And time phase is 1990 to 2000 is appointed, so its static and limited. (2).The 305days milk productions of Sanhe cattle in a farm. Type: Real; Dynamic; Unlimited. Explain: The 305days milk productions of Sanhe cattle exist ac
11、tually, so its real. And time is not defined so its dynamic and unlimited.4. (3 points each) Judge whether the following variables are likely to be continuous or discrete, if it is discrete, whether they are counting, nominal, or ordinal, and explain why please. (1).Investigating a Holstein Jersey c
12、rossbred population, recording the sex of their offspring. Type: Discrete data, nominal data Explain: The sex of the offspring can be divided into male and female.(2).The gestation length (days) in sheep carrying twins and in those carrying singletons. Type: Continuous data Explain: Within a certain
13、 range, the data can take any real value.(3).For 100 samples, each with 100 individuals, the percentages of female were recorded. Type: Discrete data; Counting data Explain: the percentages range is limited, obtained using the counting method. And the percentages are integer.(4).Lactation average lo
14、g SCC (somatic cell counts) has often been used as an indicator of CM (clinical mastitis).Type: Continuous dataExplain: Within a certain range, the data can take any real value.(5).The calving ease of goat.Type: Discrete data; Ordinal data Explain: We can change the calving ease into natural score.(
15、6).The temperature at each hour in a day. Type: Continuous dataExplain: Within a certain range, the data can take any real value, and can be measured with measurement tools. 5. (34 points) The following data are drawn from the birth height of 100 Sanhe cows ( 三河牛母牛) reared under consistent environme
16、nt in Inner Mongolia. 37373737373535353535383838383838383839393636363636363642424246464646354444444442414141414141414142424343434343433939394248484040404444474742493434343432323133334040404040404040404045454545454539393939(a) (10 points) Please divide the data into groups and calculate the mid-value
17、,frequency, cumulative frequency and relative frequency of each group.ANEWER:the number of sample is 100(1),so assume 10(912 is reasonable) groups(1),and we know the minimum is 31(1),and maximum is 49(1). The range is 18 and range of a group is 2(2). The middle of the first group is 31, and the lowe
18、r is 31 and upper is 33. The other data are drawn as the same reason.(the diagram below deserves 4)组别组中值频数累计频数频率(%)31x333233333x353469635x373613221337x393813351339x414022572241x434216731643x454412851245x474610951047x4948499449xX-XiS draw that: 29.3766Xi123.374 (1)so 137 is the outlier; (1)(2) Get ri
19、d of 137, mean=74.82 Sd=12.08 (1)Tmax=3.03 So 38.2176Xi111.42(1)So 35 is the outlier. (1)(3) Get rid of 35, mean=75.87Sd=10.30 (1)Tmax=3.01So 44.867Xi106.873(1)So 35 and 137 are the outlier. (1)7. Bonus (5 points): One farm is trying to comparing the efficiency of different culture methods, namely c
20、ulturing mussel alone (monoculture) and culturing mussels and kelp together (polyculture). Gross Weight (kg) is measured to show the culture efficiency. To do the comparison, 50 ropes of mussels were randomly selected for each culture method and Gross Weight (kg) measured when harvested, the results
21、 are as follows:Gross Weight (kg)Mussel monoculture45,45,33,53,36,45,42,43,29,25,47,50,43,49,36,30,39,44,35,38,46,51,42,38,51,45,41,51,47,44,43,46,55,55,42,27,42,35,46,53,32,41,48,50,51,46,41,34,44,46Polyculture mussels and kelp51,48,58,42,55,48,48,54,39,58,50,54,53,44,45,50,51,57,43,67,48,44,58,57,
22、46,57,50,48,41,62,51,58,48,53,47,57,51,53,48,64,52,59,55,57,48,69,52,54,53,50Please decide which method is better? You do need list all evidences to support your conclusion. (Hint: you can calculate several indicators for central tendency and dispersion tendency, to assess effect of culturing method
23、s through comparing value of those indicators)ANEWER:Average, (1)Mussel monoculture：X=1nXi=150Xi42.8Polyculture mussels and kelp：X=1nXi=150Xi52.1Range, (1)Mussel monoculture：R=max-min=55-25=30Polyculture mussels and kelp：R=max-min=69-39=30Standard deviation, (1) Mussel monoculture：S=Xi-X2n-1=7.2168P
24、olyculture mussels and kelp：S=Xi-X2n-1=6.3350Coefficient of variation, (1)Mussel monoculture：C.V.=SX100%=16.8616Polyculture mussels and kelp：C.V.=SX100%=12.1594Polyculture efficiency is better than monoculture, the reasons are as follows:1. Polyculture gross weight average is bigger than monoculture
25、, so polyculture total gross weight is bigger than monoculture; 2. Polyculture gross weight standard deviation and coefficient of variation are smaller than monoculture, so polyculture brings high level of precision than monoculture. So Mussel polyculture total weight is greater than monoculture and mussels uniformity is also higher than the monoculture. Polyculture can make better use of water, improve water quality, and increase economic efficiency. (1，答案合理即可)