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Zinaida [17]
3 years ago
7

Many unversities and colleges have instituted supplemental instruction (SI) programs, in which student facilitator meets regular

ly with a small group of students enrolled in the course to promote discussion of course material and enhance subject mastery. Suppose that students in a large statistics course are randomly divided into a control group group that will not participate in SI and a treatment group that will participate. At the end of the term each student’s total score in the course is determined.
A) Are the scores from the SI group a sample from an existing population? If so, what is it? If not, what is the relevant conceptual population?
B) What do you think is the advantage of randomly dividing the students into the two groups rather than letting each student choose which group to join?
C) Why didn’t the investigators put all students in the treatment group?
Mathematics
1 answer:
enyata [817]3 years ago
8 0

Answer:

Step-by-step explanation:

SUPPLEMENTAL INSTRUCTION PROGRAMS

- Students in a large statistics course are randomly divided into a control group and an experimental/treatment group. At the end of the term, each student's score in the course is examined

(A) The scores from the SI (Experimental) group are not a sample from an existing population. They are data from a group in a sample, hence the sample here is the number of students tested in this experiment and that equates to all the students in the large Statistics course/class.

- The relevant conceptual population would be all students in the university, since the general purpose or statement of this experiment is to see the effect of SI programs implemented in universities.

(B) The advantage of the random (unintentional) division of the students into control and experimental groups is that results or scores from the groups will be unbiased.

If you let each student choose a group, there is a high probability that the serious students will all join the SI group while the unserious ones will choose to be in the control group.

There is also a high probability that brilliant/pompous students will ignore the supplemental instruction program while students with low and medium intelligence quotient (IQ) will embrace it.

It is also possible that students will join whatever group their friends join.

These and other biases will defeat the aim of the experiment.

(C) The investigators didn't put all students in the sample into the experimental group because the difference or impact of the SI program needs to be known; whether positive or negative. Dividing them into the 2 groups lets the investigators know if there is a difference in the scores of students in the two groups.

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Step-by-step explanation:

So we have the function:

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\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta

The expression within the square root is equivalent to (Pythagorean Identity):

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta

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\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

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\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)

Distribute:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)

Now, let's make the single integral into two integrals. So:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Distribute the negative:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Divide the second and third equation by 2. So: \displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))

Therefore, our arc length will be equivalent to:

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Divide both sides by 2:

\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Evaluate:

S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))

Evaluate:

S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))

Simplify:

S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}

Use a calculator:

S\approx1.1953

And we're done!

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