Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.9 HINTS AND ANSWERS

31.9 Hints and answers

31.1 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.
31.3 (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’sthas 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.
31.5 Plot or calculate a least-squares fit of eitherx^2 versusx/yorxyversusy/x
to obtaina≈ 1 .19 andb≈ 3 .4. (a) 0.16; (b)− 0 .27. Estimate (b) is the more
accurate because, using the fact thaty(−x)=−y(x), it is effectively obtained by
interpolation rather thanextrapolation.
31.7 Recall that, because of the equal proportions of each type, theexpectednumbers
of each type in the first scheme isn. Show that the variance of the estimator for
the second scheme isσ^2 /(kn). When calculating that for the first scheme, recall
thatx^2 i=μ^2 i+σ^2 and note thatμ^2 ican be written as (μi−μ+μ)^2.
31.9 The log-likelihood function is

lnL=

∑N

i=1

lnnCxi+

∑N

i=1

xilnp+

(

Nn−

∑N

i=1

xi

)

ln(1−p);

∂(nCx)

∂n

≈ln

( n

n−x

)

−

x

2 n(n−x)

.

Ignore the second term on the RHS of the above to obtain

∑N

i=1

ln

(

n

n−xi

)

+Nln(1−p)=0.

31.11 X ̄=18. 0 ± 2 .2,Y ̄=15. 0 ± 1 .1.σˆ=4.92 givingt=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 asY^2 will not be Gaussian distributed. The difference in the means
isY ̄^2 −π^2 X ̄=47±36 and is only significantly different from zero at the 82%
level. Again the data is consistent with the proposed theory.
31.13 Consider how many entries may be chosen freely in the table if all row and
column totals are to match the observedvalues. It should be clear that for an
m×ntable the number of degrees of freedom is (m−1)(n−1).

(a) In order to make the fractions expressing each preference or lack of pref-

erence correct, the expected distribution, if there were no correlation, must

be

Classical None Pop

Mathematics 17.5 10 22.5

None 24.5 14 31.5

English 28 16 36

This gives aχ^2 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

Music preference No music preference

Academic preference 104 26

No academic preference 56 14