Chapter 15: Nonparametric Statistics
15.1
Let Y have a binomial distribution with n = 25 and p = .5. For the twotailed sign test,
the test rejects for extreme values (either too large or too small) of the test statistic whose
null distribution is the same
Chapter 13: The Analysis of Variance
13.1
2
The summary statistics are: y1 = 1.875, s12 = .6964286, y 2 = 2.625, s 2 = .8392857, and
n1 = n2 = 8. The desired test is: H0: 1 = 2 vs. Ha: 1 2, where 1, 2 represent the
mean reaction times for Stimulus 1 and 2
Chapter 14: Analysis of Categorical Data
14.1
a. H0: p1 = .41, p2 = .10, p3 = .04, p4 = .45 vs. Ha: not H0. The observed and expected
counts are:
A
B
AB
O
observed
89
18
12
81
expected 200(.41) = 82 200(.10) = 20 200(.04) = 8 200(.45) = 90
The chisquare s
Chapter 11: Linear Models and Estimation by Least Squares
Using the hint, y ( x ) = 0 + 1 x = ( y 1 x ) + 1 x = y.
11.2
a. slope = 0, intercept = 1. SSE = 6.
b. The line with a negative slope should exhibit a better fit.
c. SSE decreases when the slope ch
Chapter 12: Considerations in Designing Experiments
(
1
1 + 2
)n = ( )90 = 33.75 or 34 and n
3
3+ 5
12.1
(See Example 12.1) Let n1 =
12.2
= 90 34 = 56.
(See Ex. 12.1). If n1 = 34 and n2 = 56, then
9
25
Y1 Y2 = 34 + 56 = .7111
2
In order to achieve this s
Chapter 4: Continuous Variables and Their Probability Distributions
0.0
0.2
0.4
F(y)
0.6
0.8
1.0
4.1
y <1
0
.4 1 y < 2
a. F ( y ) = P(Y y ) = .7 2 y < 3
.9 3 y < 4
1
y4
0
1
2
b. The graph is above.
4.2
3
4
5
y
a. p(1) = .2, p(2) = (1/4)4/5 = .2, p(3) = (1
Chapter 7: Sampling Distributions and the Central Limit Theorem
7.1
a. c. Answers vary.
d. The histogram exhibits a mound shape. The sample mean should be close to 3.5 =
e. The standard deviation should be close to / 3 = 1.708/ 3 = .9860.
f. Very similar
Chapter 5: Multivariate Probability Distributions
5.1
a. The sample space S gives the possible values for Y1 and Y2:
S
AA
AB
AC
BA
BB
BC
CA
CB
CC
(y1, y2) (2, 0) (1, 1) (1, 0) (1, 1) (0, 2) (1, 0) (1, 0) (0, 1) (0, 0)
Since each sample point is equally li
Chapter 6: Functions of Random Variables
y
6.1
The distribution function of Y is FY ( y ) = 2(1 t )dt = 2 y y 2 , 0 y 1.
0
a. FU1 (u ) = P(U 1 u ) = P( 2Y 1 u ) = P(Y
u +1
2
+
+
+
) = FY ( u2 1 ) = 2( u21 ) ( u2 1 ) 2 . Thus,
fU1 (u ) = FU1 (u ) = 1u , 1
Chapter 9: Properties of Point Estimators and Methods of Estimation
9.1
Refer to Ex. 8.8 where the variances of the four estimators were calculated. Thus,
eff( 1 , 5 ) = 1/3
eff( 2 , 5 ) = 2/3
eff( 3 , 5 ) = 3/5.
9.2
a. The three estimators a unbias
Chapter 8: Estimation
8.1
Let B = B() . Then,
[
] [
]
(
)
[
2
MSE ( ) = E ( ) 2 = E ( E ( ) + B ) 2 = E E () + E ( B 2 ) + 2 B E E ()
= V ( ) + B 2 .
8.2
a. The estimator is unbiased if E( ) = . Thus, B( ) = 0.
b. E( ) = + 5.
8.3
a. Using Definition 8.3,
Chapter 1: What is Statistics?
1.1
a. Population: all generation X age US citizens (specifically, assign a 1 to those who
want to start their own business and a 0 to those who do not, so that the population is
the set of 1s and 0s). Objective: to estimate
Chapter 10: Hypothesis Testing
10.1
See Definition 10.1.
10.2
Note that Y is binomial with parameters n = 20 and p.
a. If the experimenter concludes that less than 80% of insomniacs respond to the drug
when actually the drug induces sleep in 80% of insomn
Chapter 2: Probability
2.1
A = cfw_FF, B = cfw_MM, C = cfw_MF, FM, MM. Then, AB = 0 , BC = cfw_MM, C B =
/
cfw_MF, FM, A B =cfw_FF,MM, A C = S, B C = C.
2.2
a. AB
b. A B
c. A B
d. ( A B ) ( A B )
2.3
2.4
a.
b.
8
Chapter 2: Probability
9
Instructors Soluti