Ten subjects are being studied to test the effectiveness of itch relief for three types of medication: A, B, and C. Using a 10-point scale, the researcher calculates the subjects' amount of itching before treatment compared with the amount of itching after treatment. The response variable is the itch rating after one hour of treatment. Previous studies have shown the effects of
treatment may vary based on a subject's pain resistance. How should the subjects be blocked? Explain. (4 points)
They should be blocked alphabetically so the subjects are as different as possible.
They should be blocked by age because older subjects tend to itch more.
They should be blocked by gender because female subjects tend to itch more.
They should be blocked by pain resistance so the subjects are as similar as possible.
O Blocking should not be used in this experiment.

Choice C

They should be blocked by pain resistance so the subjects are as similar as possible.

Explanation:

The process of blocking in statistics is where we group up people in blocks, and each block is determined by a characteristic of some sort. In this case, we're grouping people up by pain resistance level. Those very resistant to pain will be in one group, while those who easily feel pain will be in another block, and so on. It might be effective to have a pain scale and sort people by that.

Blocking is done so that we eliminate unwanted variability. We are testing the effectiveness of the itching cream, so all we care about is if the itch reduces or not. We aren't worried about the levels of pain. However, it says that "Previous studies have shown the effects of treatment may vary based on a subject's pain resistance". So we group up the people by pain tolerance to help eliminate the variability of pain tolerance levels.

This is why choice C is the final answer.

Extra info:

step-by-step :

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).