BASIC PRACTICE OF STATISTICS >C<
8th Edition
ISBN: 9781319220280
Author: Moore
Publisher: MAC HIGHER
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Chapter 10, Problem 10.9E
a.
To determine
To explain: Why should the review boards contain people who are not scientists?
a.
Expert Solution
Answer to Problem 10.9E
The review boards contain people who are not scientists for getting accurate reviews.
Explanation of Solution
Given info:
The institutional review boards of government regulations should consist of at least five people. One must be a scientist, one must be a non-scientist, and one must be outside the institution.
Justification:
For obtaining accurate and stable reviews of government regulations (about both the technical issue and non-technical issues of the institution) by the review boards, it must involve people who are not scientists.
Thus, the review boards contain people who are not scientists for getting accurate reviews.
b.
To determine
To explain: Whether one outside member is enough and explain how the member is chosen.
b.
Expert Solution
Answer to Problem 10.9E
Yes, one outside member is enough and a medical doctor can be chosen for giving reviews.
Explanation of Solution
Justification:
One outside member for the review board is enough as this member is selected virtually. Say, a medical doctor (who is not a scientist) would give the reviews that are not identified by the institution members.
Thus, one outside member is enough and a medical doctor can be chosen for giving reviews.
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Calculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.
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
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