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Calculate confidence intervals for
population parameters
exam
ial
Chapter 2
Confidence intervals for means of populations
notes
Essent
Covers AS 90642 (Statistics and Modelling 3.2)
Scholarship
Statistics and
Modelling
1. The fire department of Ametown claims that it will have a fire engine at the location of a fire on
average 7 minutes and 30 seconds after receiving an emergency call. It was seen in a sample of 40
house fires, that a fire engine took a mean time to turn up of 8 minutes 32 seconds. The standard
deviation was found to be 55 seconds. Use a 90% confidence interval to comment on this claim.
2. A company that sells 2 L tubs of ice-cream took a random sample of 67 such tubs. The company found
the sample had a mean of 1.85 L and a standard deviation of 0.08 L. Use a suitable confidence interval
for the true mean volume of ice-cream to comment on these findings.
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Year 2004
Ans. p. 7
Ans. p. 7
2 Scholarship Statistics and Modelling (Chapter 2)
3. During the past 12 days a postal delivery person delivered a mean of 1242 items per day with a
standard deviation of 87 items per day. What main assumption needs to be made before constructing
a 92% confidence interval?
Ans. p. 7
exam
ial
Confidence intervals for population proportions
notes
Essent
Ans. p. 7
1. A recent survey taken from a sample of motorists in the Wellington region showed that 227 people
were in favour of paying tolls to finance the construction of an alternative route out of Wellington
through Transmission Gully, and 173 were opposed.
a. Construct a 90% confidence interval for the proportion of all motorists in Wellington in favour of
paying tolls for this purpose.
b. What is the margin of error in this confidence interval?
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Calculate confidence intervals for population intervals 3
2. A survey was conducted by a hardware store, NAILS, to determine the level of customer support for a
proposed garden waste disposal service. A random sample of 200 customers was interviewed and 46
indicated their support for this proposal.
a. Find a 95% confidence interval for the percentage support of all customers for this proposal.
b. Assuming that the percentage support of all customers is the value given by the lower limit of
the interval in a, calculate the probability that the level of support from another sample of 200
customers is at least 25%.
c. NAILS claims that at least 30% of all customers support this proposal. Can this claim be justified?
Give a reason for your answer.
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Year 2004
Ans. p. 7
4 Scholarship Statistics and Modelling (Chapter 2)
Ans. p. 8
exam
ial
Confidence intervals for differences in population means
notes
Essent
d. The manager of NAILS requests that a second survey be conducted so that the width of the
interval in a is reduced by 40%. How many customers should be interviewed in the second survey?
1. A 99% confidence interval for the difference between two means is calculated from sample data. The
particular interval can be written as –4.82 < µ1 – µ2 < 9.76. What conclusions can be drawn?
Answer True or False to the following statements:
a. There is a 99% probability that the population means are equal because the interval includes 0.
b. The difference of the population means is 2.47.
c. 99% of intervals calculated in the same way will include the difference of the two means.
d. The standard error of the difference in sample means is 7.29.
Ans. p. 8
2. Samples of the times taken for male sprinters to run 100 m were taken in both 1997 and 2004. The
results are shown in the following table:
Number in sample
Mean
Standard Deviation
1997
35
10.36 seconds
0.42 seconds
2004
45
10.21 seconds
0.49 seconds
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Calculate confidence intervals for population intervals 5
Using an appropriate confidence interval, discuss the possible change in mean times for the 100 m race.
exam
ial
Finding the size of samples
notes
Essent
1. A confidence interval for a population mean has a confidence level of α and is based on a sample of
size n. The population standard deviation, σ, is known. Explain the changes that could be made to the
sample size and confidence level in order to make the confidence interval shorter.
2. A bio-genetics company is testing a variety of genetically modified potato at two different farms:
Awariki and Ruatoki. A sample of 120 potatoes from Awariki and 180 from Ruatoki is collected and
analysed for solanine content. The results, in micrograms, are shown in this table:
Awariki
Ruatoki
Mean
131.9
125.4
Standard Deviation
11.0
7.0
a. Construct a 95% confidence interval for the difference between the mean solanine contents of the
potatoes from the two farms.
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Ans. p. 8
Ans. p. 8
6 Scholarship Statistics and Modelling (Chapter 2)
b. Explain whether, on the basis of this confidence interval, it is likely that the two farms produce
potatoes with significantly different mean solanine contents.
c. Another variety of potato will be tested next year at these two farms. Assume that the standard
deviations remain the same, and that the same number (n) of potatoes is analysed from each
farm. Calculate the value of n needed to ensure that a 95% confidence interval for the difference
between the two means has a margin of error less than 2 micrograms.
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Answers
3.2 Confidence intervals for means of populations (page 1)
1. 90% confidence interval for the mean time (in seconds) of arrival, µ, of the fire trucks is:
55
55
< µ < 512 + 1.645 ×
512 – 1.645 ×
40
40
497.7 < µ < 526.3
8 min 18 s < µ < 8 min 46 s
As this interval does not contain the quoted average time of 7 minutes 30 seconds, there is insufficient
evidence at the 90% confidence level to accept the claim. The fire department should investigate
further and possibly change its arrival time estimate accordingly.
2. A 90% confidence interval for the mean capacity , µ (in L), of tubs of ice-cream is:
0.08
0.08
< µ < 1.85 + 1.645 ×
1.85 – 1.645 ×
67
67
1.83 L < µ < 1.87 L
A 99% confidence interval is:
0.08
0.08
< µ < 1.85 + 2.576 ×
1.85 – 2.576 ×
67
67
1.82 L < µ < 1.88 L
In this case it did not matter what level of confidence was used. All cases give a confidence interval
which does not include the 2 L quantity. The recommendation to the ice-cream company would be to
adjust their machinery to put more ice-cream in the containers.
3. Because n < 30, the assumption needs to be made that the number of letters being delivered each day
is Normally distributed.
3.2 Confidence intervals for population proportions (page 2)
227
400
= 400 = 0.5765
A 90% confidence interval for π is
1. a. n = 227 + 173
p=
b. Margin of error is half interval width
0.6082 – 0.5268
=
2
= 0.0407
p (1 – p )
Alternatively, margin of error= z
n
0.5675 (1− 0.5675)
400
⇒ 0.5268 < π < 0.6082
0.23 ± 1.96 ×
⇒ 0.1717 < π < 0.2883
As the question asks for percentage support then the confidence interval should be given as:
0.5675 ± 1.645 ×
46
2. a. n = 200, p =
= 0.23 , z = 1.96 (95% confidence)
200
= 0.0407
0.23 (1− 0.23)
is the 95% confidence interval for π
200
17.17% < π < 28.83%
b. This requires the Normal distribution.
p = 0.172
standard error=
0.1717 × 0.8283
200
= 0.027
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p (1 – p )
n
8 Answers and explanations
(
0.25 − 0.172
0.027
= 0.0019
)
P(support ≥ 0.25)= P Z ≥
ie level of support of at least 25% has a 0.2% probability.
c. No, as the value 30% is not in the 95% confidence interval, at the 95% confidence level there is
insufficient evidence to accept this claim.
d. Assuming the same proportion is in favour:
1.96
(
0.23 (1− 0.23)
= 0.6 × 1.96
n
1
1
= 0.62 ×
n
200
0.23(1 – 0.23)
200
)
n = 556 (to next whole number)
3.2 Confidence intervals for differences in population means (page 4)
1. a. False
b. False
c. True (on average)
d. False
2. A 90% confidence interval for the difference in means (in seconds) µ1 – µ2 is:
0.422 0.492
+
35
45
(10.36 − 10.21) ± 1.645
⇒ –0.0176 s < µ1 – µ2 < 0.3176 s
As zero is included in this interval, it can be said there is insufficient evidence at the 90% confidence
level that there is any difference between the mean times.
It is important that the conclusion drawn for these types of question is consistent with the interval
calculated
3.2 Finding the size of samples (page 5)
1. Increasing the sample size reduces the margin of error, and thus the interval is shorter. For example,
using a sample size 4 times as large (ie of size 4n) will halve the confidence interval.
2. a. A 95% confidence interval is (131.9 − 125.4) ± 1.96
112
72
+
120 180
4.28 < µ1 – µ2 < 8.72
b. As the value of zero is not included in this interval then, at the 95% confidence level, there is
a significant difference between the two means.
c. 1.96
112
72
+
<2
n
n
170
2
<
n
1.96
n > 170 ×
( )
1.96
2
2
n > 163.3
At least 164 potatoes must be sampled from each farm.
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