Practice Problems # 06 – Solutions
1. Interpolation methods are used to estimate heights above sea level for locations where direct
measurements are unavailable. In the article “Transformation of Ellipsoid Heights to Local
Leveling Heights” (M. Yanalak and O. Baykal, Journal of Surveying Engineering, 2001:90-
103), a second-order polynomial method of interpolation for estimating heights from GPS
measurements is evaluated. In a sample of 74 locations, the errors made by the method had a
sample average of 3.8 cm, with a sample standard deviation of 4.8 cm.
a. Find a 98% confidence interval for the population mean error made by this method.
b. Approximately how many locations must be sampled so that a 95% confidence interval will
specify the mean to within + 0.7 cm.
answers: Technically, we should use the Student’s T version of the confidence interval for
the mean, and that is what I use to find the answers given below. The percentage point would
be 0.01,73 2.3785t=. If you used the standard normal version of the confidence interval, then
you would use the z-percentage point, 0.01 2.32635z
=
. The resulting confidence intervals
will not be very different.
a.
() ()( )
()
0.01,73
4.8
3.8 2.3785 3.8 2.3785 0.55799
74
3.8 1.32718 2.4728 , 5.1272
s
xt n
±=± =±
=± =
The resulting interval using the standard normal version would be as follows:
() ()()
()
0.01
4.8
3.8 2.32635 3.8 2.32635 0.55799
74
3.8 1.29808 2.5019 , 5.0981
s
xz n
±=± =±
=± =
b.
()
()()
()
22
22
22 1.959964 4.8 13.43975 180.627
1.4
z
nL
α
σ
⎛⎞
⎛⎞
⎜⎟
== ==
⎜⎟
⎜⎟
⎝⎠
⎝⎠
. Hence, 181.
2. Based on tests conducted on a LARGE sample of welded joints, a 90% confidence interval for
the mean Rockwell B Hardness of a certain type of weld was computed to be (83.2 , 84.1).
Find a 95% confidence interval for the mean Rockwell B Hardness of this type of weld.
To begin, we notice that no sample size is given, but are told that the sample size is LARGE.
Hence, we can appeal to the Central Limit Theorem and use the percentiles from the
standard normal distribution, 2
z
α
. The value of the sample mean is the mid-point of this
interval; so
()
84.1 83.2 2 83.65x=+ =. One half of the interval length is the margin of
error (MOE). For this 90% confidence interval,
20.05
1.645 0.45
ss s
MOE z z
nn n
α
=== =. Thus, 0.45 0.273556
1.645
s
n== . To produce a
95% confidence interval, we use 0.025 1.95996z
=
. Hence, the confidence interval is given by:
