1. A study aims to find out whether there is a relationship between the age at which
babies begin to crawl and outside temperature at that time. Here are the data for 5 babies:
crawling age (weeks)
temperature (degrees, F)
30
66
31
73
30
72
32
63
34
48
a) (4 points) Find the mean crawling time and the mean temperature in this data set.
mean age 31.4 weeks
mean temperature 64.4 degrees
b) (4 points) The correlation between the crawling age and the temperature is
approximately r = -0.89. State whether this value of r supports of the following
statements and briefly say why for each one of them.
“There is a strong positive association between the crawling age and the outside
temperature at that time”
No, the association is strong negative, because r is negative and close to -1
“Low outside temperatures delay child’s motor development”
No, because the value of r does not say anything about one of the variables causes the
other.
c) (5 points) We would like to predict crawling age from the outside temperature.
Standard deviations for the above data are: 1.67 weeks for the age and 10.06 for the
temperature. Find the equation for the regression line, using these values together with
the values of means the correlation from parts a) and b).
Let’s call temperature x and the crawling age y.
We know that xbar=64.4, ybar=31.4
sx=10.06, sy=1.67
r=-0.89
Use formulas from the reference sheet:
b=r(sy/sx)=-0.1477
a=ybar-b*xbar=40.9119
Regression line yhat=40.9119-0.1477x
d) (2 points) Use the equation for the regression line to predict a crawling age of a child
living with the average outside temperature of 68 degrees.
yhat=40.9119-0.1477*68 = 30.71
2. Below are the numbers of home runs for 14 National League baseball teams in early
1994:
19 29 31 42 46 47 48 48 53 55 55 63 66 67
a) (3points) What is the 5-number summary of these data?
19 42 48 55 67
b) (2 points) Make a boxplot
70
60
50
40
30
20
c) (2 points) Make a stemplot
1|9
2|9
3|1
4|26788
5|256
6|367
d) (3 points) Describe the shape of the stemplot
left skewed with no outliers
3. The average lifespan of an elephant is 63 years with a standard deviation of 6 years. A
researcher makes a database of 64 elephants and records their mean lifespan.
a) (2points) What is the variable measured in this study?
lifespan of elephants
b) (3 points) What is the distribution of the recorded mean?
N(63,0.75)
c) (5 points) What is the probability that he records a number above 64?
P(xbar>64)=P((xbar-63)/.75>(64-63)/.75)=P(z>1.33)=1-0.9082=0.0918
d) (5 points) What is the lifespan duration such that the probability for the reported
number to be below that duration is .1?
Find xstar such that P(xbar<xstar)=0.1
it is the same as P(z<zstar)=0.1
from Table A zstar = -1.28
unstandardize: xstar = 63+0.75*(-1.28)=62.04
4. 0.2 is the proportion of all students like to go dancing; 0.7 is the proportion of students
who enjoy watching sports; 0.1 is the proportion of students who like to do both.
a) (2points) Draw a Venn diagram illustrating these data
.2
0.7
0.1
b) (3 points) What is the proportion of students who like either dancing or watching
sports or both?
0.1 (only dancing) + 0.6 (only sports) + 0.1 (both) = 0.8
c) (3points) What is the proportion of students who do not like to go dancing?
P(do not like dancing)=1-P(like dancing) =1-0.2 = 0.8
Statistics of the test results
Overall:
mean (percentage of the total score): 67.24
standard deviation (percentage of the total score): 24.23
five-point summary (percentage of the total score): 23 56 75 85 100
25
Number of students
20
15
10
5
0
20
30
40
50
60
70
Percentage of the total score
80
90
100
Problem 1:
30
25
Number of students
20
15
10
5
0
1
2
3
4
5
6
7
8
Score
9
10
11
12
13
14
15
Problem 2:
35
30
Number of students
25
20
15
10
5
0
1
2
3
4
5
Score
6
7
8
9
10
Problem 3:
16
14
Number of students
12
10
8
6
4
2
0
1
2
3
4
5
6
7
8
Score
9
10
11
12
13
14
15
Problem 4:
30
25
Number of students
20
15
10
5
0
0
1
2
3
4
Score
5
6
7
8