COURSE OUTLINE: Chapter R Prealgebra Review R.1 Fractions Objectives Students will be able to identify prime numbers. Students will be able to write numbers in prime factored form. Students will be able to write fractions in lowest terms. Students wi
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 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 10: Hypothesis Testing
10.1 10.2 See Definition 10.1. 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 4: Continuous Variables and Their Probability Distributions
y <1 0 .4 1 y < 2 a. F ( y ) = P(Y y ) = .7 2 y < 3 .9 3 y < 4 1 y4
1.0 F(y) 0.0 0 0.2 0.4 0.6 0.8
4.1
1
2 y
3
4
5
b. The graph is above. 4.2
a. p(1) = .2, p(2) = (1/4)4/5 = .2, p(3) = (
Chapter 16: Introduction to Bayesian Methods of Inference
16.1 Refer to Table 16.1. a. (10,30) b. n = 25 c. (10,30) , n = 25 d. Yes e. Posterior for the (1,3) prior. a.-d. Refer to Section 16.2 a.-e. Applet exercise, so answers vary. a.-d. Applex exercise
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 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.
a. Using Definition 8.3,
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 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 12: Considerations in Designing Experiments
12.1 12.2 (See Example 12.1) Let n1 =
(
1 1 + 2
)n = ( )90 = 33.75 or 34 and n
3 3+ 5
2
= 90 34 = 56.
(See Ex. 12.1). If n1 = 34 and n2 = 56, then 9 25 Y1 Y2 = 34 + 56 = .7111 In order to achieve this sa
Chapter 11: Linear Models and Estimation by Least Squares
11.1 11.2
Using the hint, y ( x ) = 0 + 1 x = ( y 1 x ) + 1 x = y.
a. slope = 0, intercept = 1. SSE = 6. b. The line with a negative slope should exhibit a better fit. c. SSE decreases when the sl
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. a. The three estimators a unbiased since:
Chapter 6: Functions of Random Variables
6.1 The distribution function of Y is FY ( y ) = 2(1 t )dt = 2 y y 2 , 0 y 1.
0 y
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 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. a. AB b. A B c. A B d. ( A B ) ( A B )
2.2 2.3
2.4
a.
b.
8
Chapter 2: Probability
9
Instructors Soluti
A set of rules that determines the order of simplifying expressions when more than one operation is involved. PEMDAS is the abbreviation used to describe the order of operations: P parentheses (or any type of grouping symbol) E - exponents M - multip
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