Exercises 6.2 page 257
6.3
A: P (-1.28 < Z < 1.28) = 79.95%
B: P (- 1.64 < Z < 1.64) = 89.90%
C:
P (-1.96 < Z < 1.96) = 5.00%
D: P (-2.33 < Z < 2.33) = 1.98%
6.4
A: P (-1.96 < Z < 1.96) = 95.00%
B:
P (-2.33 < Z < 2.33) = 98.02%
C:
P (-1.28 < Z < 1.28) = 2
Richard Surguy
MA3110
5-15-16
Exercise 4.1
Task 1
These results should not be used for 2 reasons:
- The sample sizes used are not independent; they come from a same population (authors
friends);
- The samples are not given as simple random samples and are
Richard Surguy
MA3110
5-21-16
Exercise 5.1
Task 1
Consider the samples listed in the table
A
1
3
5
B
0
6
2
5
2
C
3
12
6
3
Answers:
N
Mean
Variance
Std. Dev.
Std. Err.
1
3
9
3
35
4
2
1.15
2
5
15
3
69
6
2.45
1.10
Standard weighted Means analysis
Anova Summa
Maurice Henderson
MA3110 Statistics
Exercise 1.1
Task 1
1.1
Population- the collection of all individuals or items under consideration in a statistical study.
Sample- That part of a population form which information is obtained.
1.7
This would fall under
Richard Surguy
MA3110
4-29-16
Exercises 6.2 page 257
6.56
A: P ( Z < - 0.87 ) = 1 P ( Z < 0.87 ) = 1 0.8078 = 0.1922
B: P ( Z < 3.56 ) = 0.9998
C: P ( Z < 5.12 ) = 1.0000
6.58
A: P ( Z > 2.02 ) = 1 P ( Z < 2.02 ) = 1 0.9783 = 0.0217
B: P ( Z > -0.56 ) = P
Richard Surguy
MA3110
5-23-16
Module 6: Exercise 6.1 Linear Correlation
The obtained scatter plot is given below,
The computed correlation coefficient between these two variables is 0.8717 and the critical value
from the table is 0.754.
As the correlation
Richard Surguy
MA3110
4-10-16
Exercice 2.2
Task 1.
CBS
FOX
ABC
FOX
ABC
CBS
CBS
FOX
CBS
CBS
CBS
CBS
Media
CBS
ABC
FOX
Frequenc
y
9
5
6
20
ABC
FOX
CBS
FOX
ABC
CBS
FOX
ABC
9
8
7
6
5
4
3
2
1
0
CBS
ABC
FOX
The bar chart gives the value of the frequency for eac
Richard Surguy
MA3110
5-14-16
Exercise 3.1
Construct a 95% confidence interval estimate of the mean salary of an NCAA football coach.
Mean of 415,953
Standard deviation of 463,364.
X=
n=40
X= 415953
J= 463364
= 0.05
/2 =0.025
Z 0.025 = 1.92
E=z or/z (j/n)
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QUIZ 1
1. Find the sample standard deviation of the ages of a sample of ages of five U.S residents. 21 54 9 45 51
Answer
19.9 years
2. Data Set A had more variation than Data Set B. Decide which of the following statement is necessarily
true.
Answer
Data
MA3110: Module 5 Chi Square and Anova
What does treatment mean square (MSTR) measure?
The mean squared error of an estimator of a parameter is the expected value of the square of
the difference between the estimator and the parameter.
What does error mean
Module 5: Exercise 5.2 Chi Square Procedures
What is the population under consideration?
The Presidents of the US.
What are the two variables under consideration?
Region of birth and political party of US Presidents.
Group the bivariate data for the varia
MA3110: Module 4 Hypothesis Testing
Solve the following problems:
Examine the given statement:
1.
The proportion of people aged 1825 who currently use illicit drugs is equal to 0.20
(or 20%). Express the null hypothesis H0 and the alternative hypothesis H
MA3110: Module 4 Hypothesis Testing
Task 1: Solve the following problems:
A student of the author surveyed her friends and found that among 20 males, 4 smoke
and among 30 female friends, 6 smoke. Give two reasons why these results should not
be used for
MA3110
Linear Correlation
Exercise 6.1
By Terry Lamoreaux
The computed correlation coefficient between these two variables is 0.8717 and the critical value
from the table is 0.754.
As the correlation coefficient is more than the critical value at 0.05 sig
MA3110 Week 5 Quiz
A poll is conducted with 1434 randomly selected US adults. The poll showed that 72% of the
respondents support gun rights. The sample results are n=1434 and
. Find the
Confidence Interval that corresponds to a 95% confidence level.
a) n
MA3110
Week 5 Exercise 5.2
Chi Square and Anova
By Terry Lamoreaux
What is the population under consideration?
The Presidents of the US.
What are the two variables under consideration?
The two variables under consideration are the region of birth and poli
MA3110
MA3110
Exercise 5.1
Module 5 Chi Square and Anova
By Terry Lamoreaux
Task 1: Consider the three samples listed in the table:
A
1
3
B
0
6
5
2
5
2
C
3
1
2
6
3
Obtain the sample mean and the sample standard deviation of each of the three samples.
Samp
MA3110
Week 5
Discussion
Yes, we would be able to conduct a goodness of fit test for this experiment.
Condition 1 : The sampling method is simple random sampling.
Condition 1 is obviously true.
Condition 2 : The sample data consist of frequency counts for
MA3110
Module 4 Hypothesis Testing
Exercise 4.1
By Terry Lamoreaux
[Email address]
EXERCISE 4.1
TASK 1:
A student of the author surveyed her friends and found that among 20 males, 4 smoke and
among 30 female friends, 6 smoke. Give two reasons why these re
MA3110 QUIZ
Week 4 Quiz
1. True or False: The Mean of a sampling districution of sample means is equal to the population
mean
Answer - True
2. Assume we want the estimate the mean cost of computer repairs for the population of
consumers. How many consumer
MA3110 Statistics
Quizzes and Exams
QUIZ 3 QUESTIONS
1. A statistician wants to estimate the mean age of all college students. A random sample of 25
students indicates that the mean age is 24.6 years. Past studies showed that the standard deviation
is 1.3
MA3110
Module 3 Exercise 3.2
Sampling Distribution and Confidence Interval
By Terry Lamoreaux
Module 3: Exercise 3.2 Sample Mean Distribution and T-Interval
Task 1
Student1
Student2
Student3
Student4
Student5
Student6
Student7
Student8
Student9
Student10