Physics 2BL: Homework Set 06
Taylor Problems: 9.14, 12.3, 12.14
9.14
100 m DASH TIME (SEC): {12,11,13,14,12,15,12,16}
1500 m DASH TIME (SEC): {280, 290, 220, 260, 270, 240, 250, 230}
LET x = 100 m TIME; LET y = 1500 m TIME.
1
σ xy = ∑ ( xi − x )( yi − y )
N
σ
∑ xi yi − Nxy
r = xy =
2
2
σ xσ y
(∑ xi − Nx 2 )(∑ yi − Ny 2 )
∑ x y = 26570 sec
∑ x = 1399 sec
∑ y = 524400 sec
2
i
i
2
2
i
2
2
i
x = 13.125 sec ⇒ x 2 = 172.3 sec 2
y = 255 sec ⇒ y 2 = 65025 sec 2
r=
=
(26570 sec 2 − 8(13.125 sec)(255 sec)
[(1399 sec 2 ) − 8(172.3 sec 2 )][(524400 sec 2 ) − 8(65025 sec 2 )]
26570 sec 2 − 26775 sec 2
(20.875 sec 2 )(4200 sec 2 )
=
− 205 sec 2
87675 sec 4
= −0.692 = r
APPENDIX C: PROB8 (| r |> r0 ) ≈ 5.3%
USING A 5% SIGNIFICANCE TEST, THIS IS BORDERLINE. PERHAPS MORE
DATA IS NEEDED.
12.3
FACE SHOWING (K)
1 2 3 4 5 6
OBSERVED NUMBER (OK) 20 46 35 45 42 52
EXPECTED NUMBER (EK) 40 40 40 40 40 40
EXPECTED NUMBER FOR EACH FACE FOR UNLOADED DICE: 240/6=40
6
(O − EK ) 2
1
χ2 = ∑ K
= [(20) 2 +(6) 2 + (5) 2 + (5) 2 + (2) 2 + (12) 2 ]
EK
40
K =1
χ 2 = 15.85
SINCE THIS IS MUCH GREATER THAN n=6, THE DICE ARE LIKELY LOADED.
12.14
X
1 2 3 4 5
YOBSERVED 60 56 71 66 86
y=50+6x
56 62 68 74 80
σ =4
(a)
y − f ( xi ) ⎞
1 2 2 2 2
2
χ = ∑ ⎛⎜ i
⎟ = [4 + 6 + 3 + 8 + 6 ]
σ
⎠ 16
i =1 ⎝
χ 2 = 10.0625
5
2
2
(b) SINCE THERE IS NO CONSTRAINT FROM THE OBSERVED DATA
(MEANING THAT WE DIDN’T HAVE TO USE ANY INFORMATION ABOUT THE
OBSERVED DATA TO CALCULATE THE EXPECTED DATA), THE NUMBER OF
DEGREES OF FREEDOM IS SIMPLY EQUAL TO THE NUMBER OF DATA
POINTS, NAMELY 5.
THUS, χ~ = χ 2 / d = 10.0625 / 5 = 2.0125 .
FROM APPENDIX D, PROB( χ 2 / 5 > 2.0125) = 7.5%. > 5% .
THUS WE SHOULD NOT REJECT THE EXPECTED RELATION ON 5%
SIGNIFICANCE LEVEL.