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Q 15 Question 15 Consider the following sample of 11 length-of-stay values (measured in days): 1,...
Q 15 Question 15 Consider the following sample of 11 length-of-stay values (measured in days): 1, 1, 3, 3, 3, 4, 4, 4, 4, 5, 7 Now suppose that due to new technology you are able to reduce the length of stay at your hospital to a fraction 0.5 of the original values. Thus, your new sample is given by .5, .5, 1.5, 1.5, 1.5, 2, 2, 2, 2, 2.5, 3.5 Given that the standard error in the original sample was 0.5, in the new sample the standard error of the mean is _._. (Truncate after the first decimal.)
Solutions
Expert Solution
The standard error in the new sample = 0.5 * 0.5 = 0.25 = 2
milcah answered 23 hours agoRelated Solutions
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3
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3...
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=.05.
2
3
8
6
4
4
6
4
2
5
8
10
4
4
4
2
1
3
2
10
1
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3
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X¯= average maternity stay for a sample of
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( , ) .
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( , ) .
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P(X¯>3)=
...
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Chosen Device
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Smart Watch
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Ages
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30
80
110
40 and Up
50
40
90
Totals
80
120
200
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n
z
4 points
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Put the following wireless encryption protocols in order from
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- 1.
2. 3.
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- 1.
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Question 11 mystring.length() or mystring.size() will return the length of the string in mystring. Question 11...
Question 11
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True
False
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To access an element of an array of structures you use the
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Question 13
Question 13 options:
What would be printed from the following C++ program?
#include
#include
using namespace std;
struct Point
{
int x;
int y;
};
void makeTriangle(Point p[]);
int main()
{
Point p[3];
int i;
makeTriangle(p);...
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