Quantitative Business Methods

QBM
 Chapter 6

Estimation

Population parameters
In this lecture we estimate a population:
• Mean
• Standard deviation
• Proportion / Percentage

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Populations and Samples
• Every member of a population cannot be examined so we
use the data from a sample, taken from the same
population, to estimate some measure, such as the mean,
of the population itself.

• The sample will provide us with the best estimate of the

exact 'truth' about the population.

• The method of sampling depends on the data available

but the ideal method, as every member of the population
has an equal chance of being selected, is random
sampling.

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We estimate limits within which we are expect the 'truth'
about the population to lie and state how confident we are
about this estimation.

There are therefore two types of estimate of a population

parameter:

• Point estimate - one particular value;

• Interval estimate - an interval centred on the

point estimate

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Point Estimate of Parameter (e.g. mean)
From the sample, a single value is calculated to serve
as an estimate for the population parameter.

• The best estimate of the population

percentage, , is the sample percentage, p.

• The best estimate of the unknown

population mean, , is the sample mean.
• This estimate of  is often written û and
referred to as 'mu hat'.

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 The best estimate of the unknown population
standard deviation, , is the sample standard
deviation s, where:
(
 x−x )2

s=
(n − 1)
This is from [xσn-1] in calculator.

 (x − x )2

N.B. s = [xσn] underestimates 

(n )
Little difference between the two estimates when n is
large.
Example 1: The Accountant wishes to obtain some
information about all the invoices sent out to a
supermarket's account customers. A sample of twenty
invoices is randomly selected from the whole population
and analysed
• Values of Invoices (£):
32.53 22.27 33.38 41.47 38.05
31.47 38.00 43.16 29.05 22.20
25.27 26.78 30.97 38.07 38.06
25.11 24.11 43.48 32.93 42.04
Total = 658.44

• The proportion in the population over £40, π̂

p= 4
 100 = 20% πˆ = 20%
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• Values of Invoices (£):
32.53 22.27 33.38 41.47 38.05
31.47 38.00 43.16 29.05 22.20
25.27 26.78 30.97 38.07 38.06
25.11 24.11 43.48 32.93 42.04
Total = 658.44

• The population mean, û û = £32.92

• The population standard deviation, 658.44

x= = £32.92
• s (xn-1) = £7.12, 20
• = £7.12
Interval Estimate (Confidence interval)

• There are many applications of confidence intervals in

the field of business – see Section 6.11 of Business
Statistics for Non-Mathematicians.
• It is often more useful to quote two limits between
which the parameter, e.g. mean, is expected to lie,
together with the probability of it lying in that range.
• The limits are called the confidence limits and the
interval between them the confidence interval.
• e.g. We are 95% confident that the mean male
height lies between 5' 9" and 5' 11".

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The width of the confidence interval depends
on three sensible factors:
• the size of the sample, n;

• the amount of variation among the members

of the sample, i.e. its standard deviation, s.

• the degree of confidence we wish to have in it,

the chance of it including the 'truth', e.g. 95%;

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Standard error
• These last parameters, n and s, are used to calculate
the standard error of the mean, s/√n.

• Similarly for a percentage, p(100 − p) is

the standard error. n

• The numbers of standard errors included in the

confidence interval is found from the z- tables or t-
tables. t-tables give the wider interval needed if the
sample is only small or if we need to estimate the
standard deviation

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Summarising the choice of statistical table:

Population standard deviation

Sample
size Known: Unknown:
Standard error =σ/n Standard error = s/n

Large Normal tables: z-test Normal tables: z-test

Small Normal tables: z-test t-tables: t-test

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Percentages and Proportions
• If we know that the percentage of items with an
attribute in the sample is p and wish to form a
confidence interval for the percentage of the
items with the same attribute in the population
we use the formula:
p(100 − p)
π = pz
n
z from the normal table

• The samples must be large, ( >30), so that the

normal table may be used.
 Example 2: Out of 175 shoppers 112 preferred in-
store baked bread. Find the 95% confidence interval
for the percentage of all the store's customers who
prefer in-store baked bread.
• The point estimate for the population
112
percentage,, is p =  100 = 64%
175
• 95% Confidence Interval
p(100 − p)
• π = pz n
where p = 64 and n = 175

• 5%, two tails  z = 1.96

64  36
• p(100 − p) 64  1.96  = 64 ± 7.1
pz  175
n
• 56.9% <  < 71.1%
Confidence interval for the population mean
() where  is known (Not often the case.)
• If we actually know the population standard
deviation then we don't need to estimate it from
the sample standard deviation, i.e. we have less
uncertainty so we can be more confident.
• The normal table can therefore be used to find
the number of standard errors in the interval.

• Confidence Interval:
σ
• (z from the normal tables) μ=x z
n
Example 3:
For the whole chain of supermarkets the standard
deviation of part-time wages is known to be £1.50.
From small supermarket: n = 10, x = £6.15
• Could overall mean for small supermarket be same as for
whole chain which is £6.50?
• 95% Confidence Interval (C. I.)
1.50
6.15  1.96  = 6.15  0.930
10

• £5.22 < μ < £7.08

• Note that this interval includes £6.50 so average wages could

be the same as for the whole chain.
Confidence interval for the population mean
() where  is unknown (Usual case)
• If the sample size is small, (n < 30), and the population
standard deviation is unknown, then the t-tables are used.
• These give a wider interval and so compensates for the
probable error in estimating the value of the population
standard deviation from the sample standard deviation.

• Confidence Interval: μ = x  t s where t is

n
from the t-table with (n-1) degrees of freedom
Example 4 Continuing the Invoice example:
Find the 99% confidence interval for the mean value of all
the invoices sent out by the same small firm overleaf.
Can then mean be same as for whole chain which is
£38.50?
• x = £32.92, s = £7.12, n = 20. (from Example 1)

• Because our sample size is small, (<30), use t-tables.

• n = 20; degrees of freedom = 19; 99% C.I.; t value = 2.87

• 99% Confidence Interval

s 7.12
μ =x t  32.92  2.87 
n 20
μ = 32.92  4.57  £28.35  μ  £37.49

• Interval does not include £38.50 so mean value is not the same as
for whole chain.
Comparison of Means using 'Overlap' in
Confidence Intervals
• Can two samples have come from the same
population? We compare the intervals for the
population means.
• Do the two intervals overlap? If so, the means may
be the same. If not, they are different as they have
have no common values.
• E.g. We could compare confidence intervals for mean
spending before and after an advertising campaign.

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Example 5 The till slips of supermarket customers
were sampled both before and after an advertising
campaign and the results were analysed with the
following results:
• Before: xb = £57.60, sb = £6.70, nb = 25
• After: xa = £61.78, sa = £5.30, na = 25

• Has the advertising campaign been successful in

increasing the mean spending of all the
supermarket customers? Calculate two 95%
confidence intervals and compare the results.
• For both n = 25 so 24 d.f. , t = 2.06 for 5%, 2-tails.

• Before:
sB 6.70
μB = xB  t  57.60  2.06 
nB 25
= 57.60  2.76  £54.84  μB  £60.36

• After:
sA 5.30
μA = xA  t  61.78  2.06 
nA 25
= 61.78  2.18  £59.60  μA  £63.96
Interpretation: μB = £57.60, μA = £61.78

£54.84  μB  £60.36 £59.60  μA  £63.96

• The sample mean had risen considerably but, because the

confidence intervals overlap, the mean values for all the
sales may lie in the common ground. There may be no
difference between the two means so the advertising
campaign has not been proved to be successful.
• Note an overlap so there is a possibility of 'no difference'.
• Pre
• Post
• 55 60 65 £

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Paired Means:
requires pairs of data on each case

• The 'changes' or 'differences' are calculated and

from them their mean and standard deviation
are calculated.
sd
• Confidence Interval: μd = xd  t
nd
• x , s and n refer to the 'differences'.
d d d
Example 6 As last example but paired data used as
same cases were big spenders in both populations, i.e. the
'before's and 'after's were not independent.
• Same shoppers (A to J) were asked for till slips before and after
the advertising campaign, so the data was paired. Has there been
any mean change at a 95% confidence level?
Data £ A B C D E
Before 62.30 75.76 52.29 30.23 35.79
After 63.09 79.20 51.76 40.78 39.50

F G H I J
Before 66.50 52.30 98.65 52.20 35.90
After 70.67 57.32 97.80 57.39 37.24
First calculate the changes and then use these in the
analysis, ignoring the original data

Data £ A B C D E
Before 62.30 75.76 52.29 30.23 35.79
After 63.09 79.20 51.76 40.78 39.50
Changes 0.79 3.44 -0.53 10.55 3.71

F G H I J
Before 66.50 52.30 98.65 52.20 35.90
After 70.67 57.32 97.80 57.39 37.24
Changes 4.17 5.02 -0.85 5.19 1.34
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 From the differences:
xd = £3.28, sd = £3.37 and nd = 10
95% C.I. for the changes
sd 3.37
μd = xd  t  3.28  2.26 
nd 10

 3.28  2.41  £0.87  μ  £5.69

Note that zero is not included in the interval as both

limits have the same sign. The mean difference is always
positive so the mean spending rate has gone up after the
advertising campaign.
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Questions