Does Peyton Manning have a greater passing ability in domes? After all,
inside a dome the weather is always
nice for throwing the ball! The
accompanying relative frequency
histograms show Manning’s 18 career
road performances inside a dome and
his 86 career performances outside
(through the 2010 season). In this
case, his performance is measured
with his passer rating, a combination
or passing yards, completion
percentage, number of touchdowns,
and number of interceptions. Only
road games are included because
Manning plays his home games in a dome and the benefit of playing in a dome would be impossible to separate
from the benefit of playing at home.
a. Briefly compare these distributions.
b. State the hypotheses we are interested in testing.
c. Manning’s mean passer rating in domes was 109.3 and his mean passer rating outside was 89.8, for a
difference (dome – outside) of 19.5. Describe how to simulate the distribution of this test statistic, assuming
that Manning’s ability is the same in both locations.

d. Here are the results of 100 trials of the simulation from part c. Describe what information is provided by the
dotplot.
e. Use the dotplot to estimate and interpret the p-value.
f. Using the p-value from part e, make an appropriate conclusion.
g. Is there is convincing evidence that Manning had a greater ability in domes, can we conclude that the dome is
the cause? Explain.

did u ever get the answers for it lol. i need help on it too

ok so i had this for my stat class and i came on here to try and figure it out but didn't find any answers, so i just went ahead and tried it for myself. i'm not sure if i'm 100% correct, so if anyone knows the right answers leave a comment / post the right answers

ALSO my stat teacher gave us this exact assignment with these exact pictures so if ur gonna use these answers make sure you paraphrase them/reword them as best as you can

a) Briefly compare these distributions.

The distribution of passing ability for Manning’s passes in domes is relatively symmetrical while the distribution of his passes outside is skewed to the left. The distribution of outside performances varies slightly more than that of his performances within domes; outside performances vary from about 40-160 while performances in domes vary from about 60-160 (passer ratings).

b) State the hypotheses we are interested in testing.

H(o): Peyton Manning does not have a greater passing ability in domes than outside.

H(a): Peyton Manning has a greater passing ability in domes than outside.

c) Manning’s mean passer rating in domes was 109.3 and his mean passer rating outside was 89.8, for a difference (dome – outside) of 19.5. Describe how to simulate the distribution of this test statistic, assuming that Manning’s ability is the same in both locations.

1. Using 104 note cards, write each deviation on a different note card.

2. Shuffle the cards and divide them into two groups; one for passes outside, and one for passes within a dome.

3. Calculate and record the simulated difference (dome - outside) in standard deviation for the two types of pitches.

4. Repeat many times.

d. Here are the results of 100 trials of the simulation from part c. Describe what information is provided by the dotplot.

The dotplot illustrates the simulation repeated 100 times, with each dot representing a difference between the difference (dome - outside) for each individual simulation.

e) Use the dotplot to estimate and interpret the p-value.

Because there are no dots greater than or equal to our previously calculated difference of 19.5, the p-value = 0/100 = 0.

f) Using the p-value from part e, make an appropriate conclusion.

Since it is impossible (p = 0) to have a difference (dome - outside) in observed standard deviations of 19.5 or more by random chance alone when the true standard deviation of Manning’s passes was the same inside and outside a dome, there is convincing evidence that the true standard deviation of Manning’s passes was greater for his passes in a dome than outside a dome.

g) If there is convincing evidence that Manning had a greater ability in domes, can we conclude that the dome is the cause? Explain.

We can never 100% conclude that the dome is the primary cause of his better performance within domes. Other things like wind, crowd cheering, motivation, and teamwork could be some of the various causes of his better performance.

Step-by-step explanation:

answer: the distance

step-by-step explanation:

to calculate the distance d of a line segment with endpoints (x1, y1) and (x2, y2) use the formula d (x2 x1)2 (y2 y1)2. to calculate the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) use the formula

here we can see that the storage increases

at certain times of the year - during the

warmer months, when the gas is not much

used for heating and reserves can be

increased - and then decreases during the

colder months because the supply is being

used. it is possible to model this with a

relatively simple trigonometric graph, but

first we examine the simplest trig curves.

065/065.html

the basics

the simplest sine and cosine curves can be generated by plotting the trig-function values

against the angles used in each function. the values can be found in the chart below:

ø (˚) 0 30 45 60 90 120 135 150 180 210 225 240 270 300 315 330 360

sin ø

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0 1

2

2

2 3

2 1 3

2 2

2 1

2

0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0

cos ø

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1 3

2 2

2 1

2 0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0 1

2 2

2 3

2 1

first we plot the sine data, letting the x-axis represent the angles and the y-axis the sine

values. the result is the scatter plot below.

based on just these 17 data points, we can see

that the function is following a straightforward

pattern - the sine values are increasing to 1,

decreasing through the same values to -1, then

returning to zero. when we connect the dots, we

get the periodic curve y = sin (x) below.

this “sine curve,” with equation

y = sin(x), is the simplest of trig

graphs. the curve rises and falls by

one unit from its “axis of

oscillation” - in this case, the x-axis

- and after we cycle through 360˚ it

appears we are ready to go through

the same y-values again. we can see

this same pattern in the cosine

function on the following page.

each sine and cosine curve has two basic characteristics:

amplitude - distance the curve rises above, and falls below, its axis of oscillation (marked by

the vertical arrows on the cosine graph). the formula is

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amp = max− min

2 .

period - the length of one cycle (marked by the horizontal arrows on the cosine graph).

this can be calculated by measuring the distance between any successive maximum

points (or minimum points), or by doubling the horizontal distance between a

minimum and the next maximum.

for our first two curves, then, the amplitude is 1 and the period is 360˚ (or 2! if we are in radians).

adjusting the amplitude and frequency: y = a sin(bx), y = a cos(bx)

now that we have our “base” amplitude and frequency . . let’s change them. we can adjust both

characteristics by introducing the parameters a and b into the equations. the graph below shows the

curves y = a sin(x) for a = 1,2,3, and -0.5, with the original curve y = sin(x) being the thickest line.

clearly the amplitude of the curves has changed;

when a = 2, the amplitude is 2; when a = 3, the

amplitude is 3; and when a = -0.5 the amplitude is

0.5 (since amplitude, a distance, must be positive).

however, this last curve has also been flipped upside

down (or “reflected in the x-axis”). since the

amplitude must be positive even when a itself is

negative, we have the formula

amplitude = | a |.

the relationship between the b-value and the

period is not so straightforward. on the right

we have y = cos(bx) for b = 1,3, and 0.5. (the

thickest line is again our “base” curve.) when

b=3, the period decreases to 120˚; when

b=0.5, the period increases to 720˚ (so only

half the curve is shown here). we can describe

this reciprocal relationship with the formula

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b = 360

period or

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b = 2π

period ,

using the latter if we are working in radians.

shifting horizontally and vertically: y = sin(x-h) + k, y = cos(x-h) + k

to shift (or “translate”) a curve along either the x- or y-axis (or both), we introduce the parameters

h and k. below we have the graph of y = sin (x - 30) + 2 and y = sin(x). (note that we have returned

to an amplitude of 1 and period of 360˚). our original

curve y = sin(x) had its peak at the point (90,1); the

shifted curve has its corresponding peak at (120, 3). in

the same way, the original curve had a minimum value

at (270, -1); now the minimum is at (300,1). and

finally, the axis of oscillation has shifted vertically

from the line x = 0 to the line x =2. in other words, our

new curve has been created by shifting y = sin(x) right

30 and u p 2. and, in general, we can create the

following shifts:

horizontal shift = h (left if h < 0, right if h > 0)

vertical shift/new axis of oscillation = k.

modelling data: y = a sin[b(x-h)] + k, y = a cos[b(x-h)] + k

we have seen how the basic sine and cosine curves can be adjusted: by changing the amplitude

and period, and by shifting the curves horizontally and vertically. a summary is below.

amplitude = | a | vertical shift/axis of oscillation = k

period =

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360

b

or

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2π

b