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# Inferences: One population

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A portion of the data file for the hospital that collected the data on hand washing is shown below:

Observation Unit Time 1 Time 2
1 CCU 3 16
2 CCU 2 7
3 CCU 0 5
4 CCU 5 8
5 CCU 2 15
6 CCU 0 15
7 CCU 2 20
8 CCU 3 16
9 CCU 0 18
10 IMCU 1 16
11 IMCU 2 8

You can see that information was also collected on the unit of the employee and a second time, labeled Time 2, was also recorded. Assume that the hand-washing times are normally distributed and are recorded in seconds.

a). Find the sample means and sample standard deviations for the first hand-washing time, Time 1, for each unit.
b). Test to see if any of the units had an average hand-washing time, Time 1, less than 5 seconds.
c). The second hand-washing time, Time 2 was recorded after the employees received some training on the effects of not washing their hands long enough. Find the sample means and sample standard deviations for the second hand-washing time, Time 2, for each unit.
d). Test to see if any of the units had an average hand-washing time after training, Time 2, of less than 5 seconds.
e). Display all the initial hand-washing times, Time 1 in a histogram. Is the assumption of normality reasonable?
f). Display all the second hand-washing times, Time 2, in a histogram. Is the assumption of normality reasonable?
g). Test the hypothesis that the variance of the initial hand-washing time is 14 seconds.
h). Test the hypothesis that the variance of the hand-washing times after training is greater than 14 seconds.
i). Find the sample proportion of employees who increased their hand-washing time after training. Test the hypothesis that the proportion of employees who increased their time after training is greater than 0.80.
j). Based on your analysis, write a short report to the manager of the hospital.

https://brainmass.com/statistics/chi-squared-test/inferences-one-population-9929

#### Solution Preview

a.)
For CCU (say x):
sum(x) = 17
n = 9
mean = <x> = 17/9 = 1.89 --Answer
Variance:
=> Var(x) = sum((x-<x>)^2)/(n-1)
=> Var(x) = (sum(x^2) - n*<x>^2)/(n-1) = (55 - 9*(17/9)^2)/8 = 2.86
=> SD = sqrt(Variance) = sqrt(2.86) = 1.69 --Answer

FOR IMCU (say y):
sum(y) = 3
<y> = 3/2 = 1.5 --Answer
Variance = ((1-1.5)^2 + (2-1.5)^2)/(2-1) = (0.25+0.25)/1 = 0.5
=> SD = sqrt(variance) = sqrt(0.5) = 0.707 --Answer

b.)
with 95% confidence level: alpha = 0.05: alpha/2 = 0.025
For CCU:
mu = <x> +or- t(0.025)*SD/sqrt(9)
t(0.025) for d.f. 8 = 2.306
=> mu = 1.89 +or- 2.306*1.69/sqrt(9) = (3.189,0.591)

FOR IMCU:
t(0.025) for d.f. 1 = 12.706
mu = 1.5 +or- 12.706*0.707/sqrt(2) = (7.85, -4.85)
Units CCU has average hand-washing times less than 5 second but, IMCU can have greater than 5 sec. ...

#### Solution Summary

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