Answer Key for Homework 1 (Due date: January 24, 2003)
- Make sure to write your name and section number on each page.
- Staple everything together. Place it on the table before class starts. There is a penalty of 5 pt. if you do not staple them together. You will have to turn it in your class only.
- There is no partial credit for the numerical mistakes that you may make
- (77 pt.) Exercise 1.17
Minitab descriptive statistics for the number of transducers
Variable N Mean Median TrMean StDev SE Mean
Transducers 60 2.533 2.000 2.426 1.827 0.236
Variable Minimum Maximum Q1 Q3
Transducers 0.000 8.000 1.000 3.000
Frequency table by Minitab
Transducers Frequency Percent CumPct
0 7 11.67 11.67
1 12 20.00 31.67
2 13 21.67 53.33
3 14 23.33 76.67
4 6 10.00 86.67
5 3 5.00 91.67
6 3 5.00 96.67
7 1 1.67 98.33
8 1 1.67 100.00
n=60
Histogram by Minitab
(a)The first three columns of the frequency table above is the answer. Percent is the relative frequency.
(b)x 5 is at most5. The proportion is the relative frequency of x being 5 or less. It is also the same as the cumulative percentage for 5 transducers which is 91.67%.
x < 5 is fewer than 5. The proportion is the relative frequency of x being 4 or less. It is also the same as the cumulative percentage for 4 transducers which is 86.67%.
x 5 is at least 5. The proportion is the relative frequency of x being 5 or more which is It is also the same as the cumulative percentage for 4 transducers which is (5+5+1.67+1.67)%=13.34%.
(c) The histogram above is the answer.
(d) Identify as the data being discrete or continuous.
Discrete
(e)Calculate the average and then median number of transducers. Would you use the mean or the median to explain the center of your data? Why?
2.533
2.000
Since the data is skewed with outliers. Median should be used.
(f)Is there a mode for this data set? If there is a mode, what is it? Is this a unimodal or multimodal data?
The mode is 3. It is a unimodal data.
(g)Calculate the lower quartile, upper quartile, minimum, maximum.
Lower quartile=1
Upper quartile=3
Minimum=0
Maximum=8
(h)Calculate the range, interquartile range, variance
Range=8-0=8
IQR=3-1=2
3.3379
(i)Calculate the coefficient of variation (CV), coefficient of skewness(CS).
CV= 100(1.827/2.533)= 72.13%
CS=3(2.533-2)/1.827=0.875
(j)Construct the boxplot for this data and comment of skewness. Is the coefficient of skewness compatible with the skewness measure? Are there any outliers?
Notice that the edges of the rectangle is determined by the lower quartile,1 and the upper quartile 3. The middle line in the rectangle is determined by the median, 2. The whiskers should go as long as up to Q1-1.5(IQR)=1-1.5(2)=-2 and Q3+1.5(IQR)=3+1.5(2)=6. Since the lowest data is 0, the lower whisker stops at 0 instead of stopping at –2. The upper whisker go up to 6. The remaining observations larger than 6 are outliers.
It is positively skewed (skewness is in the direction of the longer whisker and/or outliers).
Coefficient of skewness is positive. Comments are compatible with the measure.
There are two outliers.
(k)If I collect the 61th observation as 10, would the mean, median, mode, lower quartile, upper quartile, range, interquartile range, and variance change? (If you use a spreadsheet, it will save time for you)
Variable N Mean Median TrMean StDev SE Mean
Transducers 61 2.656 2.000 2.491 2.048 0.262
Variable Minimum Maximum Q1 Q3
Transducers 0.000 10.000 1.000 3.500
The mean, upper quartile, range, interquartile range and variance change.
The median, lower quartile and mode do not change.
The minitab computes the quartiles in a different way and your hand calculations gives the upper quartile 3 whereas minitab gives it as 3.5. The upper quartile range would not change if upper quartile range does not change in this example.
- (12 pt.) Exercise 1.41. Additional to the question, answer (d) Construct the histogram for the data.
(a)7/10=0.70
(b)7/10=0.70. They are the same.
(c)x/n=0.8 then x=20 for 25 observations. 13 more S’s are necessary.
(d)
- (6 pt.) If the sample mean for 60 observations was 2.533 and the sample mean for the 25 observations was 2.320 in the same data set, what would be the sample mean for the remaining 35 observations? (You do not need the data to answer the question. Answer it using the given information here).
If the sample mean for the 60 observations was 2.533 then the sum of the 60 observations is 151.98.
If the sample mean for the 25 observations was 2.320 then the sum of the 25 observations is 58.
Then the sum of the remaining 35 observations is 151.98-58=93.98 which will give the sample mean of the remaining 35 observations as 93.98/35=2.685.
- (5 pt.) If the 3 deviations from the mean were 1.68, -2.32, -1.32 in the dataset with 60 observations. What is the sum of the remaining 57 deviations from the mean? (You do not need the data or the mean to answer the question. Answer it using the given information here).
We know that the sum of the all deviations add up to zero. Then the sum of the 3 deviations is 1.68-2.32-1.32=-1.96. It will give us the sum of the remaining 57 deviations as 0-(-1.96)=1.96.