Keith Owens
MA3110
Unit 4 Problem Set 1
Pg. 229 #2
The reasoning is correct in this statement. You cannot have half of a child and
since the number is .5 you must round up.
Pg. 229 #7This is a probability distribution with =1.5 and =0.9
x
0
1
2
3
P(x)
0.1
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Task 1
The obtained sample means and standard deviations are given below,
A
Sample Mean
Sample SD
B
3
2
C
3
2.45
6
4.24
SST = 110
SSTR = 24
SSE = 86
The constructed one way ANOVA table is given below,
ANOVA
Source of
Variation
Between Groups
W
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Week 2 Assignment 1
1. 0.22 + 0.16 + 0.21 + 0.16 = 0.75
a. Mean = N/A
b. This is not a probability distribution because the probabilities do not equal 1.00
2. 100 * 0.2 = 20
a. Mean = 20 correct answers
b. Standard deviation = 100 (0.2)(0.8)(0
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1. X= 415953 sD=463364 415953+/- (1.96)(463364)/sqrt 40= 415953+/-143598
2.
a. Acupuncture treatment : 95% CI for the mean 1.8 +/- 1.96*1.4/SQRT(142) = (1.57,
2.03)
b. 95% CI for the mean 1.6 +/- 1.96*1.2/SQRT(80) = (1.34, 1.86)
c. As the conf
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Statistics
Task 1: Understanding the Concepts and the Skills pp. 8-9
1.1 Population: The collection of all individuals or items under consideration in a statistical study.
Sample: The part of the population from which information is obtained.
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Task 1
1. The first reason why this cannot be used to claim a hypothesis test is because she used friends
instead of a random sample.
The second reason this cannot be used is because it does not satisfy the requirement of both
samples being in
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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 political party of US presidents.
Group the bivariate data
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1. Null hypothesis: H0: p = 0.20
Alternative hypothesis: H1: p 0.20
2. Since this is a two-tailed test, divide the area in half .
?/2 =0.10/2 = 0.05 area in each tail.
P (Z < z*) = 0.5
z* = -2.5, but z* also equals +2.5 because of symmetry
Z=
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The scatter plot of Wins versus Difference is shown below.
n xi yi xi
y
n x x n y y
From the given data, we have the values of n 7, x 154, y 20, x
y 1.8073, x y 118.173.
Coefficient of correlation, r
2
2
i
i
2
2
i
i
i
i
i
i
2
86, 016,
2
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Week 2 assignment 2
Task 1
Chart Title
ABC
Fox
CBS
0
0.5
1
1.5
Series 1
2
Series 2
CBS
2.5
Series 3
Fox
Series 4
ABC
3
3.5
Series 5
4
4.5
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I think the bar graph is most appropriate for showing this data. Because you can see the numb