Statistics 2
Samples and hypothesis testing
Chapter assessment
1.
A factory manager is specifying a new storage tank for a particular chemical. In
routine use, the tank will be filled to capacity each weekend. There should be
enough chemical to last until the next weekend, as emergency deliveries are very
expensive. On the other hand, money is wasted if an excessive amount of the
chemical is stored.
The volume of chemical varies from week to week and is modelled by a
Normally distributed random variable X. The manager is investigating the mean
of X. Data are available for a random sample of 15 weeks, giving the volumes of
the chemical used in each week. These are as follows (in litres).
1962
1909
1928
1940
1943
1897
1939
1924
1866
1978
1964
1944
1942
1992
1996
The standard deviation of X is taken from long experience to be 28 litres.
A 2000-litre tank will be specified if the mean of X is no more than 1930 litres.
Carry out a 5% significance test to examine whether a 2000-litre tank should be
specified, stating clearly the null and alternative hypotheses and the conclusion.
[8]
2.
A craftsman makes hand-made souvenirs. The time taken to make a souvenir is a
Normally distributed random variable with mean 34 minutes and standard
deviation 2.6 minutes.
The craftsman undertakes a training course to improve his skill. Afterwards, a
random sample of 8 times taken to make souvenirs is as follows (in minutes).
35.4
32.3
26.6
30.4
31.9
33.8
29.6
28.4
Assuming that the underlying standard deviation has not changed, test at the 0.1%
level whether the mean time taken to make a souvenir has decreased after the
training course.
[8]
3.
Psychologists are developing a new index of overall intelligence for 11-year old
children. It is assumed that the index is Normally distributed over the whole
underlying population and that the standard deviation of this distribution is 12. If
the index has been created correctly, its mean over the population should be 50.
The index is measured for a random sample of 100 11-year old children. It is
found that the sample mean value is 47.8. Test the hypothesis that the true mean
of the index is 50, against the alternative that it is not 50, at the 1% level of
significance.
[8]
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S2 Hypothesis testing Chapter assessment solutions
4.
An office experiences a lack of reliability of its email system when transmitting
messages. Many emails are successfully transmitted at the first attempt, others are
eventually successfully transmitted, but only after more than one attempt, and
others are not successfully transmitted at all. The computer manager thinks there
may be an association between the success of transmission and the type of user at
the intended destination. Results for a random sample of 300 emails are as
follows.
Type of destination
Commercial
Government
University
user
department
Successful at first
attempt
Successful after more
Transmission
than one attempt
Not successful
at all
100
57
23
21
14
13
31
21
20
(i) State the null and alternative hypotheses under examination in the usual χ²
test applied to this contingency table.
[2]
5.
(ii) Carry out the test, at the 10% significance level.
[12]
(iii) Discuss your conclusions.
[4]
As part of a survey of interest in local elections, a random sample of 845 people
was taken in towns which did not have directly-elected mayors. The people were
classified according to age (< 30 or ≥ 30) and their stated level of interest (great
or little) in local elections. The results were as follows.
Age
< 30
≥ 30
Level of interest
Great
Little
49
216
145
435
(i) Carry out the usual χ² test for independence, at the 5% significance level,
stating carefully the null and alternative hypotheses and briefly discussing
the conclusions.
[12]
At a later stage in the survey, a random sample of 1327 people was taken in
towns which had directly elected mayors. These people were classified similarly,
with the following results.
Age
< 30
≥ 30
Level of interest
Great
Little
118
314
260
635
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S2 Hypothesis testing Chapter assessment solutions
(ii) The organisation that commissioned the survey then asked whether, for
people under the age of 30, the level of interest in local elections is
independent of whether or not there is a locally elected mayor. Using the data
in the tables above, write down the 2 × 2 table, including its margins, to be
analysed.
[4]
(iii) Explain why the usual χ² test might not be appropriate for the 2 × 2 table you
have written down in part (ii).
[2]
Total 60
Solutions to Chapter assessment
1.
H0: μ = 1930
H1: μ > 1930
where μ is the population mean volume in litres
x=
29124
= 1941.6
15
EITHER: Test statistic z =
1941.6 − 1930
= 1.6045
28 15
Right-hand tail so critical value at 5% level = Φ −1(0.95 ) = 1.645
Critical region is z > 1.645
1.6045 < 1.645 so accept H0: there is not sufficient evidence to
suggests that the mean volume is greater than 1930 litres, so a
2000-litre tank should be specified.
OR:
Critical value for right-hand tail = 1930 + 1.645 ×
28
= 1941.89
15
Critical region is X > 1941.89
x is not in critical region so accept H0: there is not sufficient
evidence to suggests that the mean volume is greater than 1930
litres, so a 2000-litre tank should be specified.
OR:
⎛ 1941.6 − 1930 ⎞
P(X ≥ 1941.6) = 1 − Φ ⎜
⎟
⎝ 28 / 15
⎠
= 1 − Φ ( 1.6045 )
= 1 − 0.9457
= 0.0543
0.0543 > 0.05, so accept H0: there is not sufficient evidence to
suggests that the mean volume is greater than 1930 litres, so a
2000-litre tank should be specified.
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S2 Hypothesis testing Chapter assessment solutions
2. H0: μ = 34
H1: μ < 34
where μ is the population mean time taken in minutes
x=
248.4
= 31.05
8
EITHER: Test statistic z =
31.05 − 34
= −3.209
2.6 8
Left–hand tail so critical value at 0.1% level = Φ −1(0.001)
= −Φ −1(0.999) = −3.090
Critical region is z < -3.090
-3.209 < -3.090, so reject H0: there is evidence to suggest that the
craftsman’s mean time to make a souvenir has decreased.
OR:
Critical value for left-hand tail = 34 − 3.090 ×
2.6
= 31.16
8
Critical region is X < 31.16
x is in critical region so reject H0: there is evidence to suggest that
the craftsman’s mean time to make a souvenir has decreased.
OR:
⎛ 31.05 − 34 ⎞
P(X < 31.05) = Φ ⎜
⎟
⎝ 2.6 / 8 ⎠
= Φ ( −3.209 )
= 1 − Φ(3.209)
= 1 − 0.9993
= 0.0007
0.0007 < 0.001 so reject H0: there is evidence to suggest that the
craftsman’s mean time to make a souvenir has decreased.
3. H0: μ = 50
H1: μ ≠ 50
where μ is the population mean value of the index.
x = 47.8
47.8 − 50
= −1.8333
12 / 100
Two tail test at 1% significance so each tail is 0.5%
Critical value for left-hand tail = Φ −1(0.005 )
EITHER: Test statistic z =
= −Φ −1(0.995 ) = −2.576
Critical region is z < -2.576
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-1.8333 > -2.576 so accept H0: there is not sufficient evidence to
suggest that the true population mean is not 50.
OR:
OR:
Two tail test at 1% significance so each tail is 0.5%
12
= 46.9088
Critical value for left-hand tail = 50 − 2.576 ×
100
Critical region is X < 46.9088
x is not in critical region so accept H0: there is not sufficient
evidence to suggest that the true population mean is not 50.
⎛ 47.8 − 50 ⎞
P(X < 47.8) = Φ ⎜
⎟ = Φ( −1.8333)
⎝ 12 / 100 ⎠
= 1 − 0.9666 = 0.0334
For two tail test at 1% significance level, compare probability with
0.005.
0.0334 > 0.005
so accept H0: there is not sufficient evidence to suggest that the true
population mean is not 50.
4. (i) H0: there is no association between success of transmission and type of
destination
H1: there is some association between success of transmission and type of
destination
(ii) Observed frequencies:
Commercial
user
Successful at
first attempt
Successful after
Transmission more than one
attempt
Not successful
at all
Marginal totals
Type of destination
Government
University
department
Marginal
totals
100
57
23
180
21
14
13
48
31
21
20
72
152
92
56
300
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Expected frequencies:
Type of destination
Commercial
Government
University
user
department
Successful at
first attempt
Successful after
Transmission more than one
attempt
Not successful
at all
Marginal totals
χ
2
( 100 −
=
91.2 )
2
+
91.2
( 13 − 8.96 )
+
2
8.96
Marginal
totals
91.2
55.2
33.6
180
24.32
14.72
8.96
48
36.48
22.08
13.44
72
152
92
56
300
( 57 − 55.2 )2
55.2
( 23 − 33.6 )2 ( 21 − 24.32 )2 ( 14 − 14.72 )
+
+
+
33.6
24.32
14.72
( 31 − 36.48 )2 ( 21 − 22.08 )2 ( 20 − 13.44 )2
+
+
+
36.48
22.08
13.44
= 10.64
For a 3 × 3 table there are 2 × 2 = 4 degrees of freedom
Critical value at 10% significance level with ν = 4 4 = 7.779
10.64 > 7.770 so reject H0: there is evidence to suggest an association
between the success of the transmission and the type of destination.
(iii) When the destination is a university, far less transmissions are
successful at the first attempt than would be expected, and more are
successful after more than one attempt or unsuccessful than would be
expected. For commercial users, more are successful at the first attempt
than would be expected, and less are successful after more than one
attempt or unsuccessful than would be expected. For Government
departments the results are quite close to the expected frequencies.
5. (i) H0: there is no association between age group and level of interest
H1: there is some association between age group and level of interest
Observed frequencies
< 30
≥ 30
Marginal totals
Age
Level of interest
Great
Little
49
216
145
435
194
651
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Marginal
totals
265
580
845
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S2 Hypothesis testing Chapter assessment solutions
Expected frequencies
< 30
≥ 30
Marginal totals
Age
( 49 − 60.84 )
=
Level of interest
Great
Little
60.84
204.16
133.16
446.84
194
651
Marginal
totals
265
580
845
2
( 216 − 204.16 )2
χ
+
60.84
204.16
2
( 145 − 133.16 )
( 435 − 446.84 )2
+
+
133.16
446.84
= 4.357
For a 2 × 2 table there is 1 × 1 = 1 degree of freedom.
Critical value for 5% significance level with ν = 1 is 3.841
4.357 > 3.841
so reject H0: there is evidence to suggest an association between age group
and level of interest.
2
Under 30s seem to have less interest than would be expected, while the 30
and over age group seem to have more interest than would be expected.
(ii)
Locally
Yes
elected
No
mayor
Marginal totals
Level of interest
Great
Little
118
314
Marginal
totals
432
49
216
265
167
530
697
(iii) This is not a random sample of 697 people classified over these four cells.
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