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COMM5000 DATA LITERACY
Seminar 4 Week 5
Seminar Solutions
1. Let’s consider again the Anzac Garage data used previously. The data is in the
Excel file AnzacG.xls. Use these 117 observations on used passenger cars to find
the 95% confidence interval for the population mean distance travelled by used
passenger cars (this variable is labelled ‘odometer’ in the data set and is measured
in kilometers). Assume the population standard deviation is 60,000kms.
Since n=117 is large, we invoke the central limit theorem: �~ �, 60,0002
117
� at least
approximately. Using Excel, we find the sample mean is 78,561 kms. The 95%
confidence interval is given by
̅ ± 0.025
√
= 78,561 ± 1.96 60,000
√117 = 78,561 ± 10,87
2. What would be the effects on the width of the confidence interval calculated in the
previous question of:
(a) a decrease in the level of confidence used?
Decreases width
(b) an increase in sample size?
Decreases width
(c) an increase in the population standard deviation?
Increased width
(d) an increase in the sample standard deviation?
No effect on the width since we are told the population standard deviation.
(e) an increase in the value of ̅ found?
No effect on the width.
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3. Again, referring to the data in ‘odometer’ from AnzacG.xls and the population from
which it is drawn, determine the sample size required to estimate the population
mean to within 5,000 kms with 90% confidence. Again, assume the population
standard deviation is 60,000 kms.
ME = /2 √ ; /2= z0.05 =1.645, ME= 5,000, = 60,000 where ME is the size
of the margin of error on either side of the point estimate.
Therefore, 5,000 =
Rearranging, = �1.645×60,000
5,000 �2 = 389.67
A sample of 390 would be required.
4. A company running an urban rail service wishes to estimate its daily average
number of late-running trains on weekdays. For 10 randomly selected weekdays, it
finds the following numbers of late-running trains:
32, 10, 9, 18, 25, 15, 14, 18, 22, 16
(a) Assuming the number of late running trains on a weekday is approximately
normally distributed, calculate a 90% confidence interval for the mean
number of late-running trains on a weekday.
Let X = number of late-running trains on a weekday. Then:
= 0.1, ̅ = 17.9, 2 = 48.32, ≈ 6.9514
Since σ2 is unknown, n is small and the underlying distribution of the variable in
the population is (approximately) normal, we construct the confidence interval
using the t distribution. The required interval is:
̅
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(b) If we did not have the assumption of normality, could we still calculate a
confidence interval in this example? If not, suggest a way of overcoming this
problem.
Everything else the same, we could not construct a confidence interval in the
same way as in (a) since the t distribution is only valid if the underlying
distribution of the variable in the population is normal. This problem could be
overcome by obtaining a larger sample size and then making use of the central
limit theorem (and still using s instead of σ).
5. Reconsider the question from a previous week that used the Anzac Garage data, in
an Excel file called Anzacg.xls. Would normality be a good approximation for the
population distribution of distance travelled by used passenger cars? (Hint: look at
the summary statistics and a histogram.) Do you need to assume normality? Redo
the 95% confidence interval for the population mean distance travelled by used
passenger cars without assuming a known population standard deviation.
Excel summary statistics and histogram for distance traveled indicate non-normality. The
distribution is skewed to the right, the median is much less than the mean, and the sample
mean is only 1.35 standard deviations from zero:
Odometer (km)
Mean 78560.83
Standard
Error
5384.86
Median 67980
Mode 147000
Standard
Deviation
58246.19
Sample
Variance
3392618896
Kurtosis 3.426
Skewness 1.528
Range 315597
Minimum 403
Maximum 316000
Sum 9191617
Count 117
Frequency histogram for odometer readings for cars in
Anzac Garage data
Page 4
6. Worldwide Survey. GfK Roper surveyed people worldwide asking them “how
important is acquiring wealth to you.” Of 1535 respondents in India, 1168 said that
it was of more than average importance. In the United States of 1317 respondents,
596 said it was of more than average importance.
(a) What proportion thought acquiring wealth was of more than average
importance in each country’s sample?
1168/1535=0.761; 76.1% of Indian adults feel it is of more than average
importance.
596/1317=0.453; 45.3% of U.S. adults feel it is of more than average
importance.
(b) Create a 95% confidence interval for the proportion who thought it was of
more than average importance in India. (Be sure to test conditions.) Compare
that to the confidence interval for the U.S. population.
This is random sample of less than 10% of all adults in India; there were 1168
successes and ̂ ± 0.025��� = 0.761 ± 1.96�(0.761)(0.239)1535 = 0.761 ± 0.021 =(0.740,0.782)
We are 95% confident that between 42.6% and 48.0% of the adults in the U.S.
feel that acquiring wealth is for more than average importance. It appears that
the proportion of adults who feel this way in India is more than those in the
United States.

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