math 112

Topics: Statistics, Regression analysis, Arithmetic mean Pages: 6 (568 words) Published: May 23, 2014
STAT 208
16/05/2014
Quiz 4
1. (8-28) (18 points) An Izod impact test was performed on 20 specimens of PVC pipe. Assume that the population is normally distributed. The sample mean is ̅ and the sample standard
deviation is
. Find a 99% lower confidence bound on Izod impact strength. Sol: 99% lower confidence bound on mean Izod impact strength n  20 x  1.25 s  0.25

t0.01,19  2.539

 s 
x  t0.01,19 

 n
 0.25 
1.25  2.539 

 20 
1.108  

2. (28-5) (18 points) A random sample of 100 automobile owners shaws that, in the state of Virginia, an automobile is driven on the average 23500 kilometers per year with a standard deviation of 3900 kilometers. Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia.

Sol: 99% confidence interval, z distribution can be used
N=100 ̅
̅

̅





3. (9-58) An article in the ASCE Journal of Energy Engineering (1999, Vol. 125, pp. 59–75) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures (°C) reported were as follows: 23.01, 22.22, 22.04, 22.62, and 22.59. a. (11 points)Test the hypotheses

versus
, using
.
b. (11 points) Compute the power of the test if the true mean interior temperature is as high as 22.75.
9-54

Sol: a)
1) The parameter of interest is the true mean interior temperature life, . 2) H0 :  = 22.5
3) H1 :   22.5

4) t0 

x 
s/ n

5) Reject H0 if |t0| > t/2,n-1 where  = 0.05 and t/2,n-1 = 2.776 for n = 5 6) x  22.496 , s = 0.378, n = 5
t0 

22.496  22.5
0.378 / 5

 0.00237

7) Because –0.00237 > 2.776 we cannot reject the null hypothesis. There is not sufficient evidence to conclude that the true mean interior temperature is not equal to 22.5C at  = 0.05. b) d =

 |   0 | | 22.75  22.5 |


 0.66


0.378

Using the OC curve, Chart VII e) for  = 0.05, d = 0.66, and n = 5, we obtain   0.8 and power of 10.8 = 0.2. OR

(

)




4. (9-82) If the standard deviation of hole diameter exceeds 0.01 millimeters, there is an unacceptably high probability that the rivet will not fit. Suppose that and
millimeter.
a. (13 points) Is there strong evidence to indicate that the standard deviation of hole diameter exceeds 0.01 millimeter? Use
. State any necessary assumptions
about the underlying distribution of the data.
b. (9 points) Find the P-value for this test.
Sol: a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal.
1) The parameter of interest is the true standard deviation of the diameter,. However, the answer can be found by performing a hypothesis test on 2. 2) H0 : 2 = 0.0001
3) H1 : 2 > 0.0001
2
4)  0 =

(n  1)s 2

2

2
2
5) Reject H0 if 02   ,n 1 where  = 0.01 and 0.01,14 = 29.14 for n = 15

6) n = 15, s2 = 0.008
2
0

=

( n  1)s2
2



14(0.008)2
 8.96
0.0001

7) Because 8.96 < 29.14 fail to reject H0. There is insufficient evidence to conclude that the true standard deviation of the diameter exceeds 0.0001 at  = 0.01. P-value = P(2 > 8.96) for 14 degrees of freedom: 0.5 < P-value < 0.9 5.

(11-12) An article in the Journal of the American Ceramic Society [“Rapid Hot-Pressing of Ultrafine PSZ Powders” (1991, Vol. 74, pp. 1547–1553)] considered the microstructure of the ultrafine powder of partially stabilized zirconia as a function of temperature. The data are shown below:

a. (15 points) Fit the simple linear regression model using the method of least squares. b. (5 points) Estimate the mean porosity for a temperature of 1400 C. Sol: a)
The regression equation is
Porosity = 55.6 - 0.0342 Temperature

Predictor

Coef SE Coef

T

P

Constant

55.63 32.11 1.73 0.144...
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