Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card
12th Edition
ISBN: 9781337605199
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Brooks Cole
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Question
Chapter 7.1, Problem 19P
a.
To determine
Obtain the 90% confidence interval for the population
Find the margin of error.
a.
Expert Solution
Answer to Problem 19P
The 90% confidence interval for the population mean annual number of reported larceny cases in such communities is
The margin of error is 11.1138.
Explanation of Solution
Here,
From Table 5: Areas of a Standard
The 90% confidence interval for the population mean annual number of reported larceny cases in such communities is computed as follows:
Therefore, the 90% confidence interval for the population mean annual number of reported larceny cases in such communities is
The margin of error (E) is 11.1138.
b.
To determine
Obtain the 95% confidence interval for the population mean annual number of reported larceny cases in such communities.
Find the margin of error.
b.
Expert Solution
Answer to Problem 19P
The 95% confidence interval for the population mean annual number of reported larceny cases in such communities is
The margin of error is 13.2019.
Explanation of Solution
Here,
From Table 5: Areas of a Standard Normal Distribution, the value corresponding to 0.0250 is –1.96. That is
The 95% confidence interval for the population mean annual number of reported larceny cases in such communities is computed as follows:
Therefore, the 95% confidence interval for the population mean annual number of reported larceny cases in such communities is
The margin of error (E) is 13.2019.
c.
To determine
Obtain the 99% confidence interval for the population mean annual number of reported larceny cases in such communities.
Find the margin of error.
c.
Expert Solution
Answer to Problem 19P
The 99% confidence interval for the population mean annual number of reported larceny cases in such communities is
The margin of error is 17.378.
Explanation of Solution
Here,
From Table 5: Areas of a Standard Normal Distribution, the value corresponding to 0.0049 (which is approximately 0.005) is –2.58. That is,
The 99% confidence interval for the population mean annual number of reported larceny cases in such communities is computed as follows:
Therefore, the 99% confidence interval for the population mean annual number of reported larceny cases in such communities is
The margin of error (E) is 17.378.
d.
To determine
Compare the margin of errors and explain whether the margin of errors increases as the confidence levels increase.
d.
Expert Solution
Explanation of Solution
The margin of error obtained for 90% confidence level is 11.1138 (from Part (a)); the margin of error obtained for 95% confidence level is 13.2019 (from Part (b)); and the margin of error obtained for 99% confidence level is 17.378 (from Part (c)).
It is clear that the margin of error increases as the confidence levels increase.
e.
To determine
Compare the lengths of confidence intervals and explain whether the confidence intervals increase in length as the confidence levels increase.
e.
Expert Solution
Explanation of Solution
From Parts (a)–(c), it is clear that the length of the confidence interval for 95% confidence level is greater than the length of the confidence interval for 90% confidence level and the length of the confidence interval for 99% confidence level is greater than that for 95% and 90% confidence levels.
Therefore, the confidence interval increases in length as the confidence levels increase.
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Microsoft Excel snapshot for random sampling: Also note the formula used for the last
column
02
x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51))
A
B
1
No.
States
2
1
ALABAMA
Rand No.
0.925957526
3
2
ALASKA
0.372999976
4
3
ARIZONA
0.941323044
5
4 ARKANSAS
0.071266381
Random Sample
CALIFORNIA
NORTH CAROLINA
ARKANSAS
WASHINGTON
G7
Microsoft Excel snapshot for systematic sampling:
xfx INDEX(SD52:50551, F7)
A
B
E
F
G
1
No.
States
Rand No. Random Sample
population
50
2
1 ALABAMA
0.5296685 NEW HAMPSHIRE
sample
10
3
2 ALASKA
0.4493186 OKLAHOMA
k
5
4
3 ARIZONA
0.707914 KANSAS
5
4 ARKANSAS 0.4831379 NORTH DAKOTA
6
5 CALIFORNIA 0.7277162 INDIANA
Random Sample
Sample Name
7
6 COLORADO 0.5865002 MISSISSIPPI
8
7:ONNECTICU 0.7640596 ILLINOIS
9
8 DELAWARE 0.5783029 MISSOURI
525
10
15
INDIANA
MARYLAND
COLORADO
Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined.
1.What percent of the refunds are more than $3,500?
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A normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts?
1. Compute the probability of a value between 44.0 and 55.0.
(The question requires finding probability value between 44 and 55. Solve it in 3 steps.
In the first step, use the above formula and x = 44, calculate probability value.
In the second step repeat the first step with the only difference that x=55.
In the third step, subtract the answer of the first part from the answer of the second part.)
2. Compute the probability of a value greater than 55.0.
Use the same formula, x=55 and subtract the answer from 1.
3. Compute the probability of a value between 52.0 and 55.0.
(The question requires finding probability value between 52 and 55. Solve it in 3 steps.
In the first step, use the above formula and x = 52, calculate probability value.
In the second step repeat the first step with the only difference that x=55.
In the third step, subtract the answer of the first part from the…
Chapter 7 Solutions
Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card
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