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Question 14 Consider the following sample of 11 length-of-stay values (measured in days): 1, 1, 3,...
Question 14 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 deviation in the original sample was 1.7, in the new sample the standard deviation is _._. (Truncate after the first decimal.)
Solutions
Expert Solution
In the new sample the standard deviation is = 1.7/2 = 0.8
milcah answered 7 months agoRelated Solutions
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...
The average length of stay in U.S. hospitals 4.5 days. The length of stay in days...
The average length of stay in U.S. hospitals 4.5 days. The
length of stay in days of a random sample of patients at a local
hospital was recorded.
a. Use this data to construct a 99% confidence
interval. What are the upper and lower bounds?
b. Based on your confidence interval, is the length
of stay at this local hospital significantly different from the
national average at the 1% significance level?
Length of Stay (days)
3
6
3...
The average length of stay in U.S. hospitals 4.5 days. The length of stay in days...
The average length of stay in U.S. hospitals 4.5 days. The
length of stay in days of a random sample of patients at a local
hospital was recorded.
3,6,3,7,8,9,4,6,5,5,7,3,3,5,8,5
A). Use this data to construct a 99% confidence interval. What
are the upper and lower bounds?
B). Based on your confidence interval, is the length of stay at
this local hospital significantly different from the national
average at the 1% significance level?
The average length of a maternity stay in a U.S. hospital is 2.2 days with a...
The average length of a maternity stay in a U.S. hospital is 2.2
days with a standard deviation of 0.6 days. A sample of 25 women
who recently gave birth is taken. Find the probability that the
sample mean is greater than 2.4 days. Round your answer to the
nearest hundredth.
Given the following five pairs of (x, y) values, x 1 3 11 8 14 y...
Given the following five pairs of (x, y)
values,
x
1
3
11
8
14
y
10
7
4
2
1
(a)
Determine the least squares regression line.
(Be sure to save your unrounded values of b0
and b1 for use in Problem #6 below.)
(b)
Draw the least squares regression line accurately on a
scatterplot. Then look to see which (x, y) pairs
are above the regression line. Then add up the
y-values for all of the (x, y)...
Stay Length of Stay (LOS) Total Costs 1 3 $2,613.91 2 10 $8,769.03 3 2 $2,448.60...
Stay
Length of Stay (LOS)
Total Costs
1
3
$2,613.91
2
10
$8,769.03
3
2
$2,448.60
4
3
$2,568.70
5
3
$1,936.19
6
5
$7,230.71
7
5
$5,342.61
8
3
$4,108.13
9
1
$1,596.91
10
2
$4,061.28
11
2
$1,761.53
12
5
$4,779.19
13
1
$2,078.30
14
3
$4,713.61
15
4
$3,946.68
16
2
$2,902.74
17
1
$1,438.85
18
1
$820.21
19
1
$3,309.41
20
6
$5,476.33
INTERCEPT
Now that you have determined the answer, it is time to provide...
The length of stay (in days) from 100 randomly selected hospital patients are presented in the...
The length of stay (in days) from 100 randomly selected
hospital patients are presented in the table below. Suppose we want
to test the hypothesis that the population mean length of stay at
the hospital is less than 5 days. Conduct hypothesis using
=.05.
2
3
8
6
4
4
6
4
2
5
8
10
4
4
4
2
1
3
2
10
1
3
2
3
4
3
5
2
4
1
2
9
1
7
17...
The length of maternity stay in U.S. hospitals is normally distributed, with mean of 3.5 days...
The length of maternity stay in U.S. hospitals is normally
distributed, with mean of 3.5 days and a
standard deviation of 1.2 days. We randomly
survey 91 women who recently bore children
in U.S. hospitals.
Let X= maternity stay in U.S. hospital, and
X¯= average maternity stay for a sample of
size 91 .
X∼ (pick one) BPEN
( , ) .
X¯∼ (pick one) EPNB
( , ) .
What is the probability that the average stay is more
than 3 days?
P(X¯>3)=
...
Consider the following daily time series for days 1-14. Use this information to answer the questions...
Consider the following daily time series for days 1-14. Use this
information to answer the questions below.
Day
Daily Steps
1
2270
2
3869
3
3285
4
4598
5
3539
6
2474
7
2246
8
4800
9
3696
10
4817
11
3261
12
1839
13
5121
14
2800
Step 1 of 7: What is the 3-day Moving Average forecast for time
period 4?
Round to the nearest whole number.
Step 2 of 7: What is the 3-day Moving Average forecast...
Question 1 (of 11)Question 2 (of 11)Question 3 (of 11)Question 4 (of 11)Questions 5 - 6...
Question 1 (of 11)Question 2 (of 11)Question 3 (of 11)Question 4
(of 11)Questions 5 - 6 (of 11)Questions 7 - 9 (of 11)Questions 10 -
11 (of 11) Save & ExitSubmit
Time remaining: 0:51:30
Problem 7-5A Determine depreciation under three methods
(LO7-4)
[The following information applies to the questions
displayed below.]
University Car Wash built a deluxe car wash across the street
from campus. The new machines cost $270,000 including installation.
The company estimates that the equipment will have a residual...
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