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NARA [144]
3 years ago
15

In 2000, the mayo clinic conducted a study of whether air dryers eliminate more germs than paper towels. researchers recruited 1

00 peop,e contaminated their hands, and then instructed them to wash with soap and water for one minute. afterward, they randomly assigned 50 participants to run their hands under a warm air dryer for a single 30 second cycle and 50 participants to use a paper towel for 15 seconds. the methods did not produce a significant difference in the mean bacterial counts. what type of study is this?
Mathematics
1 answer:
WARRIOR [948]3 years ago
3 0

Answer:

This is an observational study based on a stratified sample.

Step-by-step explanation:

This study aimed to check which clean-up/drying up method after handwashing was more effective; that is, do dryers eliminate more germs than paper towels.

It involved 100 people that took part in the study. The 100 of them contaminated their hands and were instructed to wash their hands with soap and water for 15 seconds before the participants are randomly grouped into two strata; one stratum/group (consisting of 50 participants) that dries their hands with a dryer and another (also consisting of 50 participants) that dry their hands with paper towels.

An observational study is one in which data is collected only by monitoring the events of the study. The conductor of the study in no way tries to intervene or control any of the variables being studied. No control, no tweaking of factors, just observe and record data.

An observational study based on a stratified sample is an observational study that uses the stratified sampling method to carry out its objectives. It randomly puts each of the participants being observed into groups or strata. Each group of randomly assigned participants are then required to run the test for one of the variables being compared.

This is exactly what the study in the question describes. This is why this study is undoubtedly an observational study based on a stratified sample.

Hope this Helps!!!

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LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

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Answer:

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