STATS:DATA+MODELS(LL)-W/ACCESS>CUSTOM<
21st Edition
ISBN: 9780137643219
Author: DeVeaux
Publisher: PEARSON
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Question
Chapter 16, Problem 1E
a.
To determine
Explain the shape of the sampling distribution for the sample proportion.
a.
Expert Solution
Answer to Problem 1E
The shape of the sampling distribution for the sample proportion is unimodal and symmetric.
Explanation of Solution
Given info:
A random sample of 200 is drawn from customers who recently logged in and checked their IP address and the smartphone users true proportion is 36%.
Justification:
Sampling distribution for a proportion:
For all possible samples the distribution of proportion is called the sampling distribution of proportion
- Samples should be independent.
Sample size should be large.- The sampling distribution of
mean p and standard deviation of
Here random sample is drawn by considering 200 investors as a small portion of their customer. Therefore, the sampling distribution for the proportion of 200 investors that use smartphones would be unimodal and symmetrical.
b.
To determine
Find the mean of this sampling distribution.
b.
Expert Solution
Answer to Problem 1E
The mean of this sampling distribution is 0.36.
Explanation of Solution
Justification:
From part (a), the mean for the sampling distribution of proportion is p.
Here, the true proportion smartphone users, is 36% that is, 0.36.
Thus, the mean of this sampling distribution is 0.36.
c.
To determine
Find the standard deviation of the sampling distribution.
c.
Expert Solution
Answer to Problem 1E
The standard deviation of the sampling distribution is 0.034.
Explanation of Solution
Calculation:
From part (a), the standard deviation of
Here proportion p is 0.36 and q is
Substitute p as 0.36, q as 0.64 and n as 200 in the formula,
Thus, the standard deviation of the sampling distribution is 0.034.
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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
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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.
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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.)
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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 16 Solutions
STATS:DATA+MODELS(LL)-W/ACCESS>CUSTOM<
Ch. 16.2 - 1. You want to poll a random sample of 100...Ch. 16.2 - Prob. 2JCCh. 16.2 - Prob. 3JCCh. 16.3 - Prob. 4JCCh. 16.3 - Prob. 5JCCh. 16.3 - Prob. 6JCCh. 16.3 - Prob. 7JCCh. 16.3 - Prob. 8JCCh. 16.5 - Prob. 9JCCh. 16.5 - Prob. 10JC
Ch. 16.5 - Think some more about the 95% confidence interval...Ch. 16 - Prob. 1ECh. 16 - 2. Marketing The proportion of adult women in the...Ch. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - 5. Living online Pew Research, in 2015, polled a...Ch. 16 - 6. How’s life? Gallup regularly conducts a poll...Ch. 16 - 7. Marriage According to a Pew Research survey,...Ch. 16 - 8. Campus sample For her final project, Stacy...Ch. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - 12. Soup A machine is supposed to fill cans with...Ch. 16 - Prob. 13ECh. 16 - 14. Spanking In a 2015 Pew Research study on...Ch. 16 - Prob. 15ECh. 16 - 16. Smoking The Gallup poll described in Exercise...Ch. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - 20. Hiring In preparing a report on the economy,...Ch. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - 24. More conditions Consider each situation...Ch. 16 - Prob. 25ECh. 16 - 26. More conclusions In January 2002, two students...Ch. 16 - Prob. 27ECh. 16 - 28. Confidence intervals, again Several factors...Ch. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - 31. Mislabeled seafood In 2013 the environmental...Ch. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - 34. Still living online The Pew Research poll...Ch. 16 - Prob. 35ECh. 16 - 36. Take the offer First USA, a major credit card...Ch. 16 - Prob. 37ECh. 16 - 38. Junk mail Direct mail advertisers send...Ch. 16 - Prob. 39ECh. 16 - 40. Local news The mayor of a small city has...Ch. 16 - Prob. 41ECh. 16 - 42. Gambling A city ballot includes a local...Ch. 16 - Prob. 43ECh. 16 - 44. Teachers A 2011 Gallup poll found that 76% of...Ch. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - 48. Back to campus II Suppose ACT, Inc. wants to...Ch. 16 - Prob. 49ECh. 16 - Prob. 50ECh. 16 - Prob. 51ECh. 16 - Prob. 52ECh. 16 - Prob. 53ECh. 16 - Prob. 54E
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