Statistics 101 – Homework 2
Solution
Reading:
January 23 – January 25
January 28 –
Chapter 4
Chapter 5
Assignment:
1. As part of a physiology study participants had their heart rate (beats per minute)
taken by a trained nurse practitioner. Below are heart rates for a sample of 20
females.
70
66
71
76
73
78
71
62
69
77
75
72
77
61
57
74
68
79
68
75
a) Answer the questions, Who? What? for these data.
Who? Females.
What? Heart rate (beats per minute).
b) Make a stem-and-leaf display of the females’ heart rates (use a split stem).
5 |
5*| 7
6 |21
6*| 9 8 8 6
7 |013124
7*| 5 7 6 8 7 9 5
5 |
5*| 7
6 |12
6*| 6 8 8 9
7 |011234
7*| 5 5 6 7 7 8 9
c) Using the stem-and-leaf display, describe the distribution of heart rates for
females. Make sure you mention the shape, center, spread and any outliers or
other interesting characteristics of the distribution.
The shape is mounded on the high side (on the right) and skewed towards
lower values (to the left).
The center is in the low 70’s beats per minute.
The spread is from 57 to 79 beats per minute.
There are no apparent outliers or other unusual characteristics.
d) Calculate the sample mean heart rate.
The sample mean heart rate is 70.95 beats per minute.
1
e) Calculate the sample median heart rate. Why is the sample mean heart rate
lower than the sample median heart rate?
The sample median heart rate is 71.5 beats per minute. The sample mean
is lower than the sample median because the distribution is skewed to the
left and the sample mean tends to be pulled in the direction of the skew.
f) Calculate the five number summary for these data.
minimum:
lower quartile:
median:
upper quartile:
maximum:
g) Given the value
for these data.
s=
57.0 bpm
68.0 bpm
71.5 bpm
75.5 bpm
79.0 bpm
∑ ( y − y ) = 680.95, calculate the sample standard deviation
2
680.95
= 35.83947 = 6.0 beats per minute
19
h) Which do you think provides a more informative summary of these data; a
five number summary or the sample mean and sample standard deviation?
Briefly explain your answer.
The shape of the distribution is skewed left. A five number summary
would be a more informative summary.
The heart rates for a sample of 20 males were also taken.
i) Describe the distribution of heart rates for males. Make sure you mention the
shape, center, spread and any outliers or other interesting characteristics of the
distribution.
The shape of the distribution is symmetric and mounded in the middle.
The center is between 70 and 75 beats per minute.
The spread is from 64 to 82 beats per minute.
There are no apparent outliers or other unusual characteristics.
j) Give the value of the sample mean heart rate. Give the value of the sample
median heart rate.
Sample mean heart rate is 72.35 beats per minute.
Sample median heart rate is 72 beats per minute.
2
k) Give the value of the sample standard deviation for male heart rate.
The sample standard deviation is 4.7 beats per minute.
l) Construct side-by-side box plots (use a common scale) and use these to
compare the distribution of females’ and males’ heart rates.
85
Heart Rate
80
75
70
65
60
55
F
M
Gender
Males’ heart rate is slightly higher on average than females. Females also
show more variation, or spread, than males do. It is hard to see the left
skew shape in the box plot for females but the symmetric box plot for
males confirms our earlier observation.
2. JMP Assignment: What is a “normal” body temperature? Researches took the body
temperature (degrees Fahrenheit) of 130 adults, 65 males and 65 females. The data
are in a file on the course web page. Follow the instructions in the JMP Guide to
download the data from the web and open the data in JMP.
a) Obtain a histogram, box plot, stem-and-leaf display and descriptive statistics
for the body temperature. Print the JMP output and turn it in with your
assignment. Use the output to answer the following questions.
• Describe the distribution of body temperature. Make sure to mention the
shape, center, spread and any outliers.
The shape of the distribution of body temperature is fairly symmetric.
The center is around 98.3 degrees Fahrenheit.
The data is spread from 96.3 to 100.8 degrees Fahrenheit.
The box plot indicates three potential outliers; two on the low end (96.3
and 96.4 degrees F) and one on the high end (100.8 degrees F)
3
•
What percentage of body temperatures are less than 98.6 degrees
Fahrenheit?
81 out of 130 or 62.3% of the body temperatures are less than 98.6
degrees Fahrenheit.
•
What is the sample median body temperature? What is the sample mean
body temperature?
The sample median body temperature is 98.3 degrees Fahrenheit.
The sample mean body temperature is 98.25 degrees Fahrenheit.
•
Comment on the relationship between the sample median and sample
mean and how this relationship is consistent with your description of the
shape of the distribution.
The sample median and the sample mean are almost equal. This is
consistent with the description of the shape of the distribution as
symmetric.
b) Obtain histograms and descriptive statistics for the body temperature by
gender. Make sure histograms have common scales. Also obtain side-by-side
box plots for males and females. Print the JMP output and turn it in with
your assignment. Use the output to answer the following questions.
• Compare males and females in terms of body temperature.
Females tend to have slightly higher body temperatures than males. The
sample mean and sample median for females are 98.39 oF and 90.40 oF,
respectively while the sample mean and median for males is 98.10 oF.
Both distributions are fairly symmetric. There are a few potential
outliers for the females, the temperature of 100.8 oF is unusually high
while temperatures of 96.4, 96.7 and 96.8 oF are unusually low. Looking
at all of the values, the females have a larger standard deviation, 0.7435
o
F, than the males, 0.6988 oF. Ignoring these potential outliers, the two
distributions have about the same spread.
4
Distribution of Body Temperature for a sample of 130 adults.
40
20
Count
30
10
96
97
98
99
100
101
Body Temperature (F)
Stem
100
100
99
99
98
98
97
97
96
96
Leaf
8
0
59
000001112223344
555666666666677777777888888888899
00000000000111222222222233333444444444
556666777888888899999
0111222344444
7789
34
Count
1
1
2
15
33
38
21
13
4
2
96|3 represents 96.3
Quantiles
100.0%
75.0%
50.0%
25.0%
0.0%
maximum
quartile
median
quartile
minimum
Moments
100.80
98.70
98.30
97.80
96.30
Mean
Std Dev
N
98.249231
0.7331832
130
5
Gender=F (Female)
Gender=M (Male)
20
10
5
5
96
97
98
99
100
Count
10
15
Count
15
101
97
98
99
100
101
Quantiles
100.0%
75.0%
50.0%
25.0%
0.0%
maximum
quartile
median
quartile
minimum
100.80
98.80
98.40
98.00
96.40
100.0%
75.0%
50.0%
25.0%
0.0%
maximum
quartile
median
quartile
minimum
99.500
98.600
98.100
97.600
96.300
Moments
Mean
Std Dev
N
98.393846 Mean
0.7434878 Std Dev
65 N
98.104615
0.6987558
65
101
Temp
100
99
98
97
96
F
M
Gender
Means and Std Deviations
Level
Number
Mean
F
65
98.3938
M
65
98.1046
.
Std Dev
0.743488
0.698756
6