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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