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alisha [4.7K]
3 years ago
11

A university was interested in examining the overall effectiveness of its online statistics course, along with the effectiveness

of particular aspects of the course. First, the university wanted to see whether the online course was better than a standard course. Second, the university wanted to know whether students learned best using Excel, using Minitab, or using no statistical package at all. The university randomly selected a group of 30 students and administered one of the different variants of the course (i.e., traditional or online, coupled with one of the three software options) to each student. The success of each variant was measured by the students' average improvement between a pre-test and a post-test.
i. Is this a controlled experiment or observational study?
ii. What are the explanatory variables?
A. The students mean pre-test average score.
B. Whether the course was online or traditional.
C. Whether the course was online or traditional and the statistical software used.
iii. The response variable is:
A. the students' mean post-test score.
B. the difference between the students' mean pre-test and post-test scores.
C. an evaluation of course quality provided by an independent rater.
iv. How many treatment groups are there in this study?
A. 3
B. 4
C. 5
D. 6
E. 7
Mathematics
1 answer:
kodGreya [7K]3 years ago
5 0

Answer:

i. This is controlled experiment

ii. C. Whether the course was online or traditional and the statistical software used.

iii. B. The difference between students mean pre-test and post-test scores.

iv. D. 6

Step-by-step explanation:

This is controlled experiment because the university is observing students who take either online course or a standard course. The students have 3 different software available which they can learn. The students are observed based o pre test and post test results.

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6 0
3 years ago
A)A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹.
Nutka1998 [239]

Answer: (given assumed typo corrections)


(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.


The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.



Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.


Step-by-step explanation:

This problem has typos. Assuming:

Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.


129.6 cm^2 = 2(base cm^2) + 4(side cm^2)

= 2(X cm)^2 + 4(X cm)(2X cm)

= (2X^2 + 8X^2)cm^2

= 10X^2 cm^2

X^2 cm^2 = 129.6/10 = 12.96 cm^2

X cm = √12.96 cm = 3.6 cm


so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,

X(t) = 0.01t + 3.6

= the length parameter after t seconds, in cm.


V(X) = 2X^3 cm^3

= the volume when the length parameter is X.


dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt

that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule


V'(X) = 6X^2 cm^3/cm

= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).


X'(t) = 0.01 cm/sec

= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3

= the volume after t seconds, in cm^3

V'(X(t)) = 6(0.01t+3.6)^2 cm^2

= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec

= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.


Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?


Question part b)


Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.


This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).


Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.

Then we want y'(x+Δx/2)×Δx


y(x) = 2x^2 + 3x

y'(x) = 4x + 3


y(4) = 44

y(4.02) = 44.3808

Δy = 0.3808

Δy/Δx = (0.3808)/(0.02) = 19.04


y'(4) = 19

y'(4.01) = 19.04

y'(4.02) = 19.08


Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.


y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.


In this case, where y is quadratic in x, this method gives Δy exactly.

6 0
3 years ago
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