interval carbonate

```09.10.25
L&iacute;kindareikingur og t&ouml;lfr&aelig;&eth;i
Vor&ouml;nn 2004
&AElig;fing &iacute; &ouml;ryggisbilum
&Aacute;kve&eth;i&eth; hva&eth;a tilfelli (sj&aacute; samantekt um &ouml;ryggisbil) &aacute; vi&eth; hvert d&aelig;mi h&eacute;r fyrir ne&eth;an.
1.
Let 1 and 2 denote true average tread lives for two competing brands of size FR7815 radial tires. Compute a 95% confidence interval for the difference 1 and 2 when
the observed data on the first brand yield a sample mean of 36,500 hours and a sample
standard deviation of 2,200 hours (sample size n = 40) while the observed data on the
second brand yield a sample mean of 23,800 hours and a sample standard deviation of
1,900 hours (sample size m = 40). Assume that the two populations are normal and
that the variances are equal.
2.
A random sample of n = 8 E-glass fiber test specimens of a certain type yielded a
sample average interfacial shear yield stress of 30.2 and a sample standard deviation
of 3.1. Assuming that interfacial shear yield stress is normally distributed, compute a
95% confidence interval for true average stress.
3.
The accompanying table summarizes some data from an experiment carried out to
study various characteristics of anchor bolts.
Bolt diameter (in.)
Samle size
Sample mean shear
Sample standard
strength (kip)
deviation
3/8
78
4.25
1.30
1/2
88
7.14
1.68
Let 1 and 2 be the true average shear strength for bolts with 3/8 and 1/2 in.
diameter, respectively. Calculate a 95% confidence interval for the difference 1 - 2.
Assume that the two populations are normal and that the variances are not equal.
4.
State DMW records indicate that if all vehicles undergoing emissions testing during
the previous year, 70% passed the first try. A random sample of 200 cars tested in a
particular county during the current year yields 156 that passed on the initial test.
Calculate a 90% confidence interval for the true proportion of cars that passed on the
initial test and check if the interval includes 0.70.
5.
The amount of lateral expansion (mils) was determined for a sample of n = 9 pulsedpower gas metal arc welds used in LNG ship containments tanks. The resulting
sample variance was s2 = 7.90. Assuming normality, derive a 95% confidence interval
for 2.
6.
For each of 18 preserved cores from oil-wet carbonate reservoirs, the amount of
residual gas saturation after a solvent injection was measured at water flood-out. The
observed mean and standard deviation are 38.7 and 10.8, respectively. Compute a
98% confidence interval for the true average amount of residual gas saturation.
7.
Trace metals in drinking water affect the flavor, and unusually high concentrations
can pose a health hazard. A certain paper reports on a study in which six river
locations were selected and the zinc concentration (mg/L) determined for both surface
water and bottom water at each location. The six pairs of observations are displayed
in the accompanying table. Compute a 95% confidence interval for the mean
difference in zinc concentration between surface water and bottom water. Assume
normality.
Location
Zinc
1
2
3
4
5
6
0.430
0.266
0.567
0.531
0.707
0.716
0.415
0.238
0.390
0.410
0.605
0.609
0.015
0.028
0.177
0.121
0.102
0.107
conc. in
bottom
water
Zinc
conc. in
surface
water
Differenc
e
8.
A random sample of n = 15 heat pumps of a certain type yielded the following
observation on lifetime (in years): 2.0, 1.3, 6.0, 1.9, 5.1, 0.4, 1.0, 5.3, 15.7, 0.7, 4.8,
0.9, 12.2, 5.3, 0.6. Assume that the lifetime distribution is exponential. Compute a
95% confidence interval for the expected lifetime.
9.
An article in the Los Angeles Times (Dec. 12, 1977) reports that the technique
graphic stress telethermometry (GST) accurately detected 23 out of 29 known breast
cancer cases. Construct a 90% confidence interval for the true proportion of breast
cancers that would be detected by the GST technique.
10.
A paper on relative controllability of dissimilar cars reports results of an experiment
to compare handling ability for two cars having quite different lengths, wheelbases,
and turning radii. The observations are time in seconds required for subjects to
parallel park each car. Compute a 95% confidence interval for the mean difference in
time to park. Assume normality.
Subject
1
2
3
4
5
6
7
Car A
37.0
25.8
16.2
24.2
22.0
33.4
23.8
Car B
17.8
20.2
16.8
41.4
21.4
38.4
16.8
Subject
8
9
10
11
12
13
14
Car A
37.0
25.8
16.2
24.2
22.0
33.4
23.8
Car B
17.8
20.2
16.8
41.4
21.4
38.4
16.8
11.
An experiment to compare the tension bond strength of polymer latex modified
mortar to that of unmodified mortar resulted in x = 18.12 kgf/cm2 for the modified
mortar (n = 40) and y = 16.87 kgf/cm2 for the unmodified mortar (m = 32). Let 1
and 2 be the true average tension bond strengths for the modified and unmodified
mortars, respectively. Compute a 99% confidence interval for the difference 1 - 2.
The standard deviation of the two population are known, 1 = 1.6 (modified mortar)
and 2 = 1.4 (unmodified mortar). Assume that the two populations are normal.
12.
Observations were made on fracture toughness of base plate of 18% nickel maraging
steel (in ksi in.). Calculate a 99% confidence interval for the standard deviation of
the fracture toughness distribution. The observed standard deviation is 5.21, based on
n = 22 observations. Assuming the data are normally distributed, calculate a 99%
confidence interval for the standard deviation of the fracture toughness distribution.
13.
A sample of 300 urban adults residents of a particular state revealed 63 who favored
increasing the highway speed limit from 55 to 65 mph, while a sample of 180 rural
residents yielded 75 who favored the increase. Compute a 95% confidence interval for
the difference between the proportion of urban residents and the proportion of rural
residents that favored the increase.
14.
A 95% confidence interval is desired for the true average stray-load loss  (watts) for
a certain type of induction motor where the line current is held at 10 amps for a speed
of 1500 rpm. Assume that stray-load loss is normally distributed with known  = 3.0.
The sample mean is 58.3 and is based on 25 observations.
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