Biostatistics

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1. Each year, the National Highway Traffic Safety Administration (NHTSA) crash tests new car models
to determine how well they protect the driver and the front-seat passenger in a head-on collision.
The NHTSA has developed a “star” scoring system for the frontal crash tests, with results ranging from
one star (*) to five stars (*****). The more stars in the rating, the better the level of crash protection
in a head-on collision. The NHTSA crash test results for 98 cars in a recent model year are given below.
Star Score Freq.
2 4
3 17
4 59
5 18
Part A: Use the frequency distributed data to calculate the descriptive statistics below.
You may want to use the descriptive statistics for frequency data worksheet.
N =
Mean=
Sample Variance=
Standard Deviation=
Standard Error=
CV=
Skewness (g1) =
Kurtosis (g2) =
Part B Use the data above to draw a frequency histogram (bar chart). Be sure to label the axes
appropriately and use good graphing technique. When you are finished, move it into the space
provided below. You can use the corners of the plot to adjust the size if necessary.
2. A study of the characteristics of cheek teeth (e.g., molars) in an extinct primate species was published
in a recent (2010) issue of the American Journal of Physical Anthropology. Data on dentary depth of
molars (in millimeters) for 18 cheek teeth extracted from skulls are reproduced below.
Dentary Depth (mm)
18.12
19.48
19.36
15.94
15.83
19.70
15.76
17.00
13.96
16.55
15.70
17.83
13.25
16.12
18.13
14.02
14.04
16.20
Calculate the statistics listed below for the data on dentary depth. Pay attention to decimal places and be
sure to include the units. You may use any method (cell functions or the descriptive statistics data
analysis tool) to obtain your answers.
Units
Mean=
Median=
Mode=
Sample Variance=
Standard Deviation=
Standard Error=
Kurtosis=
Skewness=
95% Confidence Limits Units
Upper Boundary=
Lower Boundary=
3. The American College of Obstetricians and Gynecologists reports that 32% of all births in the US take
place by Cesarean section each year. (National Vital Statistics Reports, 2010). If, at a local Denver
Hospital, 25 women gave birth last weekend.
What’s the expected number of women who gave birth by Cesarean section?
Answer =
What's the probability that the number of Cesarean section births was 11?
Answer =
What's the probability that the number of Cesarean sections was fewer than 4?
Answer =
What's the probability that the number of Cesarean sections was 10 or more?
Answer =
4. A “planet transit” is a rare celestial event in which a planet appears to cross in front of its star as
seen from Earth. The planet transit causes a noticeable dip in the star’s brightness, allowing scientists
to detect a new planet even though it is not directly visible. The National Aeronautics and Space
Administration (NASA) recently launched its Kepler mission, designed to discover new planets in
the Milky Way by detecting extrasolar planet transits. After one year of the mission in which 3000
stars were monitored, NASA announced that five planet transits were detected. Assume the number of
transits discovered for every 3000 stars follows a Poisson distribution with a mean of 5. What’s the
probability that, in the next 3000 stars monitored by the Kepler mission:
more than 10 planet transits will be seen?
Answer =
exactly 3 planet transits will be seen?
Answer =
fewer than 5 planet transits will be seen?
Answer =
5. Aquatic Biology (2010) reported on a study of green sea turtles inhabiting the Grand Cayman South
Sound Lagoon. The data on curved carapace (shell) length (in centimeters) for 23 captured turtles
are given below. Do the appropriate statistical test to answer the question of whether or not the
sample of turtles belongs to a population with a mean of 52.6 cm.
Carapace Length (cm)
42.43
49.96
46.04
48.76
47.78
45.81
49.05
49.65
49.71
54.29
52.01
51.15
54.42
52.62
53.27
54.07
50.40
53.69
51.30
54.29
54.58
55.11
57.65
Ho =
Ha =
test statistic =
df =
Exact probability of the test statistic =
Conclusion relative to the hypothesis.
6. Because they share an identical genotype, twins make ideal subjects for investigating the degree
to which various environmental conditions affect personality. The classical method for studying this
phenomenon is the study of identical twins separated early in life and reared apart. The data below
represent IQ scores for 32 pairs of identical twins where one twin (A) was reared by a natural parent
and the other twin (B) was reared by a relative or some other person. Do the appropriate statistical
test to determine if there is a significant difference between the average IQ scores of identical
twins when one member of the pair is reared by the natural parents and the other member of the
pair is not. Make no assumptions and show all work.
Pair ID TWIN-A TWIN-B
112 113 109
114 94 100
126 99 86
132 77 80
136 81 95
148 91 106
170 111 117
172 104 107
174 85 85
180 66 84
184 111 125
186 51 66
202 109 108
216 122 121
218 97 98
220 82 94
228 100 88
232 100 104
236 93 84
306 99 95
308 109 98
312 95 100
314 75 86
324 104 103
328 73 78
330 88 99
338 92 111
342 108 110
344 88 83
350 90 82
352 79 76
416 97 98
Ho =
Ha =
ts =
df =
The exact probability of ts =
Conclusion relative to the hypothesis.
7. Do a correlation analysis on the IQ scores for the identical twins in problem 6.
Put your statistical results below.
Ho =
Ha =
test statistic =
df =
Exact probability of the test statistic =
Conclusion relative to the hypothesis.
8. A dietitian has developed a diet that is low in fats, carbohydrates, and cholesterol. Although the
diet was initially intended to be used by people with heart disease, the dietitian wishes to examine
the effect of this diet on the weights of obese people. Two random samples of 25 obese people are
selected, and one group of 25 is placed on the low-fat diet (low-fat). The other 25 are placed on a diet
that contains approximately the same quantity of food, but is not as low in fats, carbohydrates, and
cholesterol (regular). For each person, the amount of weight lost (in LBS) in a three-week period was
recorded and presented below. Do the appropriate statistical test to determine whether the
weight loss on the low-fat diet was greater than the weight loss on the regular diet. Make no
assumptions and show all work.
Low-Fat Regular
8 2
10 6
10 11
12 7
9 9
3 8
11 5
7 8
9 7
2 6
21 2
8 6
9 8
2 5
2 7
20 6
14 8
11 6
15 8
6 13
13 1
8 9
10 8
12 12
1 10
Ho =
Ha =
test statistic =
df =
Exact probability of the test statistic =
Conclusion relative to the hypothesis.
What is the Statistical Power of this test?:
Answer: %
Statistical Power Calculator: https://www.dssresearch.com/KnowledgeCenter/toolkitcalculators/statisticalpowercalculators.aspx
9. In all-electric homes, the amount of electricity expended is of interest to consumers, builders, and
groups involved with energy conservation. Suppose we wish to investigate the monthly electrical
usage in all-electric homes and its relationship to size of the home. The data are given below.
Monthly Usage
Home Size (ft2) kilowatt-hours
1290 1182
1350 1172
1470 1264
1600 1493
1710 1571
1840 1711
1980 1804
2230 1840
2400 1956
2710 2007
2930 1984
3000 1960
Part A. Is there a significant functional relationship between electricity utilization and the size of the home?
In other words, does monthly electricity usage depend on home size? Do the appropriate statistical test
and put your results in the boxes provided.
Ho =
Ha =
test statistic =
df =
Exact probability of the test statistic =
Conclusion relative to the hypothesis.
Part B. Using appropriate graphic skills, plot the data presented in Part A. Include the
regression line and the r2 value in your plot. Put the plot below. Make the plot attractive
and use good graphing techniques (pay attention to size of axes labels, colors, etc)
10. The data below are from a published report on children who repeat a grade in elementary school
(Archives of Disease in Childhood, 2004). The researchers compared Australian schoolchildren who
repeated a grade and recorded the data for girls and boys separately. Do the appropriate statistical
test to determine whether repeating a grade is independent of gender.
Boys Girls
Never Repeated a Grade 1349 1366
Repeated Grade 86 43
Ho =
Ha =
test statistic =
df =
Exact prob. of the test statistic =
Conclusion relative to the hypothesis.
11A. Pediatric researchers at Pennsylvania State University carried out a designed study to test whether
a teaspoon of honey before bed calms a child’s cough (Archives of Pediatrics and Adolescent Medicine,
2007). A sample of 63 children, who were ill with an upper respiratory tract infection, and their parents
participated in the study. On the first night, the parents rated their children’s cough symptoms on a
scale of 0 (no problems at all) to 30 (extremely severe). On the second night, the parents were
instructed to give their sick child a dosage of liquid “medicine” prior to bedtime. Unknown to the
parents, some were given a dosage of dextromethorphan (DM), an over-the-counter cough medicine,
while others were given a similar dose of honey. A third group of parents (the control group) gave
their sick children no dosage at all. Again, the parents rated their children’s cough symptoms, and
the improvement in total cough symptoms score was determined for each child. Is there evidence
that the treatments were significantly different with respect to improvement score?
Honey DM Control
10 3 7
6 4 7
10 9 12
8 12 7
11 7 9
12 6 7
12 8 9
8 12 5
12 12 11
9 4 9
11 12 5
15 13 6
10 7 8
15 10 8
9 13 6
13 9 7
8 4 10
12 4 9
10 10 4
8 15 8
9 9 7
Ho =
Ha =
test statistic =
df =
Exact probability of the test statistic =
Conclusion relative to the hypothesis.
11B. Which means in the data set above are significantly different from each other?
Perform a Tukey's HSD Procedure on the data above in order to determine the
answer to this question. The online calculator can be found at:
http://astatsa.com/OneWay_Anova_with_TukeyHSD/
Copy and paste the online results out to the below.
Using the lines provide below, underline means that are not significantly different at the .05 level
Use the minimum number of lines that are necessary to accurately depict the relationships between the means.
The length of a line can be changed by clicking on it and then grabbing either end with the mouse cursor.
Honey DM Control
10.4 8.7 7.7
12. In the Journal of Personality and Social Psychology, psychologists investigated the potential harmful
effects of violent music lyrics. The researchers theorized that listening to a song with violent lyrics
will lead to more violent thoughts and actions. A total of 60 undergraduate college students participated
in an experiment designed by the researchers. Half of the students were volunteers, and half were
required to participate as part of their introductory psychology class. Each student listened to a
song by the group “Tool,” with half the students randomly assigned to a song with violent lyrics
and half assigned a song with nonviolent lyrics. After listening to the song, each student was given
a list of word pairs and asked to rate the similarity of each word in the pair on a seven-point scale.
One word in each pair was aggressive in meeting (e. g., choke) and the other was ambiguous (e. g.,
night). An aggressive cognition score was assigned on the basis of the average word-pair scores. The
higher the score, the more the subject associated an ambiguous word with a violent word. The data
are given below. Conduct the appropriate ANOVA on these data and put your results in the box
provided.
Psychology
Class Volunteer
Non-violent Lyrics 2.5 2.4
2.9 2.4
2.9 2.5
3.0 2.6
2.6 3.6
2.4 4.0
3.5 3.3
3.3 3.7
3.7 2.8
3.3 2.9
2.8 3.2
2.5 2.5
2.8 2.9
2.0 3.0
3.1 2.4
Violent Lyrics 3.4 4.1
3.9 3.5
4.2 3.4
3.2 4.1
4.3 3.7
3.3 2.8
3.1 3.4
3.2 4.0
3.8 2.5
3.1 3.0
3.8 3.4
4.1 3.5
3.3 3.2
3.8 3.1
4.5 3.6
Analyze these data using the appropriate model of 2-way ANOVA. Give your results using standard format
(the hypotheses and your conclusions relative to the hypotheses).
13. With respect to the data presented in problem 12, what if the data on violent versus non-violent
lyrics were taken from four randomly selected introductory psychology class sections (out of the
eight sections of introductory psychology classes that were available). So, instead of psychology
class versus volunteer in the columns, each of the four data groups were obtained from the four
randomly selected sections. Two of the sections listened to a “Tool” song with non-violent lyrics
and two of the sections listened to a song with violent lyrics. Since the class sections are not replicated
within the lyric treatments, this represents a nested ANOVA design, with class sections nested within
lyric treatments.
Class Sections
Section A Section B
Non-violent Lyrics 2.5 2.4
2.9 2.4
2.9 2.5
3.0 2.6
2.6 3.6
2.4 4.0
3.5 3.3
3.3 3.7
3.7 2.8
3.3 2.9
2.8 3.2
2.5 2.5
2.8 2.9
2.0 3.0
3.1 2.4
Violent Lyrics 3.4 4.1
3.9 3.5
4.2 3.4
3.2 4.1
4.3 3.7
3.3 2.8
3.1 3.4
3.2 4.0
3.8 2.5
3.1 3.0
3.8 3.4
4.1 3.5
3.3 3.2
3.8 3.1
4.5 3.6
Analyze the data above as a 2-level nested ANOVA. Put your hypotheses and your conclusions in the box provided
using our standard format. The correct ANOVA table should be put out to the right. Think about which
factor is the top level in this analysis.
Hypotheses:
Conclusions:
Enjoy the rest of your summer!
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