Hints and answers

31.9 HINTS AND ANSWERS
31.9 Hints and answers
31.1
31.3
31.5
31.7
31.9
Note that the reading of 9.28 m s−2 is clearly in error, and should not be used in
the calculation; 9.80 ± 0.02 m s−2 .
(a) 55.1. (b) Note that two thirds of the readings lie within ±2 of the mean and
that 14 readings are being used. This gives a standard error in the mean ≈ 0.6.
(c) Student’s t has a value of about 2.5 for 13 d.o.f. (degrees of freedom), and
therefore it is likely at the 3% significance level that the data are in conflict with
the accepted value.
Plot or calculate a least-squares fit of either x2 versus x/y or xy versus y/x
to obtain a ≈ 1.19 and b ≈ 3.4. (a) 0.16; (b) −0.27. Estimate (b) is the more
accurate because, using the fact that y(−x) = −y(x), it is effectively obtained by
interpolation rather than extrapolation.
Recall that, because of the equal proportions of each type, the expected numbers
of each type in the first scheme is n. Show that the variance of the estimator for
the second scheme is σ 2 /(kn). When calculating that for the first scheme, recall
that x2i = µ2i + σ 2 and note that µ2i can be written as (µi − µ + µ)2 .
The log-likelihood function is
N
N
N
n
ln L =
ln Cxi +
xi ln p + Nn −
xi ln(1 − p);
i=1
i=1
i=1
n x
∂(n Cx )
≈ ln
−
.
∂n
n−x
2n(n − x)
Ignore the second term on the RHS of the above to obtain
N
n
+ N ln(1 − p) = 0.
ln
n − xi
i=1
31.11
31.13
X̄ = 18.0 ± 2.2, Ȳ = 15.0 ± 1.1. σ̂ = 4.92 giving t = 1.21 for 14 d.o.f., and
is significant only at the 75% level. Thus there is no significant disagreement
between the data and the theory. For the second theory, only the mean values
can be tested as Y 2 will not be Gaussian distributed. The difference in the means
is Ȳ 2 − π 2 X̄ = 47 ± 36 and is only significantly different from zero at the 82%
level. Again the data is consistent with the proposed theory.
Consider how many entries may be chosen freely in the table if all row and
column totals are to match the observed values. It should be clear that for an
m × n table the number of degrees of freedom is (m − 1)(n − 1).
(a) In order to make the fractions expressing each preference or lack of preference correct, the expected distribution, if there were no correlation, must
be
Classical
17.5
24.5
28
Mathematics
None
English
None
10
14
16
Pop
22.5
31.5
36
This gives a χ2 of 12.3 for four d.o.f., making it less than 2% likely, that no
correlation exists.
(b) The expected distribution, if there were no correlation, is
Academic preference
No academic preference
Music preference
104
56
1303
No music preference
26
14
STATISTICS
This gives a χ2 of 1.2 for one d.o.f and no evidence for the claim.
31.15
As the distribution at each value of s is Poisson, the best estimate
of the
√
measurement error is the square root of the number of counts, i.e. n(s). Linear
regression gives a = 4.3 ± 2.1 and b = 10.06 ± 0.94.
(a) The cosmic ray background must be present, since n(0) = 0 but its value of
about 4 is uncertain to within a factor 2.
(b) The correlation coefficient between a and b is −0.63. Yes; if a were reduced
towards zero then b would have to be increased to compensate.
(c) Yes, χ2 = 4.9 for five d.o.f., which is almost exactly the ‘expected’ value,
neither too good nor too bad.
31.17
31.19
a1 = 2.02 ± 0.06, a2 = −2.99 ± 0.09, a3 = 4.90 ± 0.10; r12 = −0.60.
Note that |dF| = |dF /F 2 | and write
N1 − 1
N1 − 1
(N2 − 1)F .
1+
as
1
+
(N2 − 1)F (N2 − 1)F N1 − 1
1304