Happiness comes not from material wealth but less desire
1
Inferences Comparing Two
Population Means
Chapter
6
Confidence intervals
Statistical tests
Sample size selection
3
4
Estimation for
Point estimator
Confidence interval
1.
2.
Normal populations wi
STAT 6304 HOMEWORK SIX
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STAT
Stat. 6304 (Fall, 2011) HW 1 Solutions
6.1 No, the sample mean does not have to correspond to a data point. For example, suppose the data set
is cfw_0, 1. The sample mean is 0.5, which is not part of the data set.
6.2 No. Using the Jordan data set from cl
Stat. 6304 (Fall, 2011) HW 2 Solutions
3.3 The possible values are cfw_0, 1, 2, ., 99999.
3.4 The possible values are cfw_0, 1, 2, ., 10.
3.24 The probability mass function is
0.3
p(x) = 0.5
0.2
0
if x=$10,000,000
if x=$5,000,000
if x = $1, 000, 000
othe
Stat. 6304 (Fall, 2011) HW 3 Solutions
1. P (a X b) = P (a < X b) + P (X = a) = P (a < X b) + 0 = P (a < X b). P (a < X b) =
P (a < X < b) + P (X = b)P (a < X < b) + 0 = P (a < X < b). P (a X b) = P (a X < b) + P (X =
b) = P (a X < b) + 0 = P (a X < b).
2
Homework 4 Solutions
2.5-1
a. This is Binomial
with n = 1 and p = 2/3. Thus, = 2/3, 2 = 2/9
q
and = 29
b. Thisq
is Binomial with n = 12 and p = 3/4. Thus, = 9, 2 = 9/4 and
= 94
2.5-5
This is an example of the Geometric distribution with p = 1/5.
P (X = 4
Homework 5 Solutions
3.5-1
a. Here, X Gamma( = 10, = 3/2), so the p.d.f. is
f (x) =
b.
M (t) =
x9 e2x/3
(10)( 23 )10
1
1
=
,
(1 t)
(1 (3/2)t)10
t < 2/3
X = = 10(3/2) = 15
2
X
= 2 = 10(3/2)2 = 45/2
3.5-8
Here, X P ois( = 16).
a. W Gamma( = 7, = 1/16).
b. H
Stat. 6304 (Fall, 2011) HW 6 Solutions
9-99. This problem can be done in MINITAB, using STAT > Tables > Chi-Square goodness of fit (1
Var). Put the observed data in a column and the probabilities below (the probabilities X equals the
given values, except
Stat. 6304 (Fall, 2011) HW 8 Solutions
10-55. Use Stat > Basic Statistics > 2 Variances for this problem.
a. H0 : men = women . Ha : men 6= women . = 0.02. Assuming that the individual values in
both groups are normal, F = 0.92 and the p-value = 0.843. Th
Homework 2 Solutions
5.6-1
Assuming n = 12 is large enough for the CLT to apply,
1
N 1/2,
.
X
12 12
Therefore
2/3) = P
P (1/2 X
1/2 1/2
2/3 1/2
Z
1/12
1/12
= P (0 Z 2) = 0.4472.
5.6-8
= E(X) = 24.43
a. E(X)
= V ar(X)/n = 2.20/30 = 0.0733
b. V ar(X)
c.
Homework 3 Solutions
6.1-1
Let X1 , X2 , . . . , Xn be a random sample from the N (, 2 ), where
= cfw_ : < < .
Then the likelihood function is given by
P
(xi )2
2 n/2
L() = (2 )
exp
,
2 2
.
Then, the log-likelihood function is
P
2
`() = (n/2) ln(2 )
Homework 1 Solutions
5.1-4
Parts (c) and (d) will vary.
a.
0, x 0
x2 , 0 < x < 1
FX (x) =
1, x 1
b. We can simulate X by setting Y = FX (X) = X 2 X = Y .
We can then generate random U(0, 1) realizations and plug them in for
Y in the above to get realizat
Stat. 6304 (Fall, 2011) Homework Assignment 3
1. (Problem 4-12) If X is a continuous random variable, argue that P (a < X b) = P (a < X b) =
P (a X < b) = P (a < X < b).
2. (Problem 4-28) Suppose f (x) = x/8 for 0 < x < 4.
a. Show that f (x) is a probabil
Homework 5 Solutions
6.3-12
For this data, xd = 0.07875 and sd = 0.2549222.
a. A point estimate of d is xd = 0.07875.
b. t.05;23 = 1.714, so the confidence interval is (0.01043932, ).
c. Since the lower bound is negative, there is not enough evidence to s
Homework 6 Solutions
6.5-1
a. p = 0.03738318
b. (0.02270901, 0.05205734)
c. (0.02524793, 0.0550219)
d. (0.02504435, 0.055451)
p
e. p + 1.645 p(1 p)/642 = 0.04969899.
6.5-3
a. p = 690/1000 = .69.
b. There are varying acceptable answers depending on which f
Chapter 6: Inferences Comparing Two Population Central
Values
6.2 Inferences about 1 2 Independent Samples
6.1
a. Reject Ho if |t| 2.064
b. Reject Ho if t 2.624
c. Reject Ho if t 1.860
6.2 Ho : 1 2 0 versus Ha : 1 2 < 0; Reject Ho if t 1.703
71.579.8
1 1
Chapter 11: Linear Regression and Correlation
11.2 Estimating Model Parameters
40
45
11.1 A scatterplot of the data is given here:
35
30
20
y
25
5
10
15
5
10
15
20
x
11.2
a. y = 1.8 + 2.0(3) = 7.8
b. The equation is plotted here:
160
25
12
11
10
9
8
7
6
0
When we free ourselves of
desire,
we will know serenity and
freedom.
1
Inferences about Population
Means
Sections
5.1 -5.7
Estimation
Statistical tests: z test and t test
Sample size selection
2
Estimation
To estimate a numerical summary in population
(pa
Since everything is a reflection of our minds,
everything can be changed by our minds.
1
Random Variables
Section
4.6-4.14
Types of random variables
Binomial and Normal distributions
Sampling distributions and Central
limit theorem
Random sampling
Normal
Since everything is a reflection of our minds,
everything can be changed by our minds.
1
Inferences about Population
Variances
Sections
7.1 to 7.3
The distribution of s^2
Tests for equal variances
2
The Distribution of S^2
Let Y1, Y2, , Yn be a random sam
Always be mindful of the kindness and not the faults of others.
1
One-way Anova: Inferences about
More than Two Population Means
Model
and test for oneway anova
Assumption checking
Nonparamateric
alternative
2
Analysis of Variance & One Factor
Designs (On
Learn to let go. That is the key to happiness.
~Jack Kornfield
1
Probability
Section
4.1-4.5
Basic terms and rules
Conditional probability and
independence
Bayes rule
2
This lady has lost 10
games in a row on
this slot machine.
Would you play this
slot ma
Simple Linear Regression
Linear
regression model
Prediction
Limitation
Correlation
1
Example: Computer Repair
A company markets and repairs small computers.
How fast (Time) an electronic component
(Computer Unit) can be repaired is very important
to the e
Always be mindful of the kindness and not the faults of others.
1
Categorical Data
Sections
10.1 to 10.5
Estimation for proportions
Tests for proportions
Chi-square tests
Example
Researchers
in the development of new
treatments for cancer patients often
Normal, Binomial, Poisson Distribution/CLT/Normality Checking
(R Textbook Section 3.5)
Four fundamental items can be calculated for a statistical distribution:
Density or point probability (starting with d)
Cumulated probability, distribution function (st
Introduction to R Commander
For people who lack of programming experiences, you may need a crutch to help you learn R
smoothly. R Commander is a basic graphical user interface for R. You can create graphics
without typing a line code, purely point-and-cli
Getting Started With R
Installation.
The R software package can be obtained free from www.r-project.org. To install R on a
Windows machine go to this web address; in the left margin under Download, select CRAN;
select the CRAN Mirror site nearest you; sel
Getting Started With the R Commander
John Fox and Milan Bouchet-Valat
Version 2.3-0 (last modified: 2016-08-09)
1
Introduction
The R Commander (Fox, 2005, 2017) provides a graphical user interface (GUI) to the open-source
R statistical computing environme
STAT 6304 HOMEWORK THREE
4.36
a. Continuous
b. Discrete
c. Continuous
d. Discrete
e. Continuous
f. Continuous
4.44
No, each trial does not result in one of two outcomes.
1|Page
STAT 6304 HOMEWORK THREE
2|Page
STAT 6304 HOMEWORK THREE
4.76
a. Use a compute