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chubhunter [2.5K]
3 years ago
12

Mice and Pain Can you tell if a mouse is in pain by looking at its facial expression? A new study believes you can. The study1 c

reated a "mouse grimace scale" and tested to see if there was a positive correlation between scores on that scale and the degree and duration of pain (based on injections of a weak and mildly painful solution.) The study's authors believe that if the scale applies to other mammals as well, it could help veterinarians test how well painkillers and other medications work in animals. 1"Of Mice and Pain", The Week, May 28, 2010, p. 21. 1. State the null and alternative hypotheses. 2.Since the study authors report that you can tell if a mouse is in pain by looking at its facial expression, do you think the data were found to be statistically significant? 3.If another study were conducted testing the correlation between scores on the "mouse grimace scale" and a placebo (non-painful) solution, should we expect to see a sample correlation as extreme as that found in the original study? 4. If another study were conducted testing the correlation between scores on the "mouse grimace scale" and a placebo (non-painful) solution, should we expect to see a sample correlation as extreme as that found in the original study, if the original study results showed no evidence of a relationship between mouse grimaces and pain?
Mathematics
1 answer:
kotykmax [81]3 years ago
3 0

<u>Explanation:</u>

1. a) Null hypothesis: There is <em>no</em> statistically significant relationship between the mouse grimace scale and the amount of pain felt by mouse.

b) Alternate hypothesis: There is a statistically-significant relationship between the mouse grimace scale and the amount of pain felt by mouse.

2. Yes, because a statistically significant data implies that there is sufficient evidence to believe the study, based on the results of the findings.

3. No, since the variables are different in this case. Here we are dealing with a non-painful solution so there may be no sample correlation as extreme as that found in the original study.

4. Possibly, because every hypothesis is an assumption until it is proven. Thus, in every statistical research, there may be different findings.

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Find the distance between P1(4,16degrees) and P2(-2,177degrees) on the polar plane.
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You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.

Let's convert (4, 16°) to Cartesian coordinates.

Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.

cosine = adjacent / hypotenuse
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sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327<span>

So (4, 16°) </span>⇒ (3.84504678375, 1.10254942327).

Let's convert (-2, 177°)  to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)

In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).

Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.

cosine = adjacent / hypotenuse
cosθ = x/r
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sine = opposite / hypotenuse
sinθ = y/r
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So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).

Now we must use the distance formula with our two points.
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}
d\approx\sqrt{-1.84778771^2+-1.20722134^2}
d\approx\sqrt{3.41431942+1.45738336}
d\approx\sqrt{4.87170278}
\boxed{d\approx2.20719342}

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