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PERSISTANT LOW HEMOCRIT IN ALL THE ELDERLY PATIENTS IN A NURSING HOME F .31, WITH A NORMAL OF .41. AN INTERVENTION PROGRAM OF AN IRON SUPPLEMENT WAS INITIATED TO SEE IF THE HEMOCRIT WOULD INCREASE.
A RANDOM SAMPLE OF 100 PATIENTS SHOWED THAT 60 PATIENTS NOW HAD ELEVATED LEVELS OF HEMOCRIT OF .41.
CAN WE CONCLUDE AT ALPHA= 0.05 THAT THE INTERVENTION WAS A SUCCESS?
a. what is the error value.
b. state the null and alternative hypotheses
c. define the test statistic
d. develope the decision rule.
e.compute the test statistic
f. state the conclusion.

Solution Preview

HI there,
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<br>First we sould clarify the problem.
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<br>Normally - the frequency of hemocrit is 0.31 - therfore - 100 people, there should be 31 who have hemocrit. (Normal hemocrit levels are 0.41).
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<br>Now an experiment is done to test if giving iron will increase the hemocrit level of seniors.
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<br>SO the expected number of seniors who have elevated hemocrit is 31/100
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<br>When 100 seniors were given iron, now we see that 60 had elevated hemocrit.
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<br>In this problem, we are going to compute a test statistic called the T-test. BUT we have a major assumption here. In this problem, the stanard ...

Solution Summary

The solution addresses a PERSISTANT LOW HEMOCRIT IN ALL THE ELDERLY PATIENTS IN A NURSING HOME F .31, WITH A NORMAL OF .41. AN INTERVENTION PROGRAM OF AN IRON SUPPLEMENT WAS INITIATED TO SEE IF THE HEMOCRIT WOULD INCREASE.
A RANDOM SAMPLE OF 100 PATIENTS SHOWED THAT 60 PATIENTS NOW HAD ELEVATED LEVELS OF HEMOCRIT OF .41.
CAN WE CONCLUDE AT ALPHA= 0.05 THAT THE INTERVENTION WAS A SUCCESS?
a. what is the error value.
b. state the null and alternative hypotheses
c. define the test statistic
d. develope the decision rule.
e.compute the test statistic
f. state the conclusion.

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