math 112 Essay

STAT 208

16/05/2014

Quiz 4

1 . (8-28) (18 points) A great Izod effects test was performed about 20 individuals of PVC pipe. Imagine the population is usually distributed. The sample indicate is М… and the sample standard

change is

. Discover a 99% decrease confidence bound on Izod impact strength. Sol: 00% lower self-confidence bound on mean Izod impact strength n пЂЅ 20 x пЂЅ 1 . 25 h пЂЅ 0. 25

t0. 01, 19 пЂЅ installment payments on your 539

пѓ¦ s пѓ¶

x пЂ­ t0. 01, 19 пѓ§

пѓ·п‚ЈпЃ­

пѓЁ nпѓё

пѓ¦ 0. twenty-five пѓ¶

1 ) 25 пЂ­ 2 . 539 пѓ§

пѓ·п‚ЈпЃ­

пѓЁ twenty пѓё

1 . 108 п‚Ј пЃ­

installment payments on your (28-5) (18 points) A random test of 95 automobile owners shaws that, in the express of Virginia, an automobile is driven around the average 23500 kilometers each year with a common deviation of 3900 miles. Construct a 99% assurance interval to get the average number of kilometers a vehicle is motivated annually in Virginia.

Encanto: 99% self confidence interval, unces distribution works extremely well

N=100 М…

М…

М…

в€љ

в€љ

3. (9-58) An article in the ASCE Log of Energy Anatomist (1999, Volume. 125, pp. 59–75) identifies a study of the thermal masse properties of autoclaved oxygenated concrete employed as a building material. Five samples of the material were analyzed in a framework, and the typical interior conditions (°C) reported were as follows: 23. 01, 22. 22, 22. '04, 22. 62, and twenty-two. 59. a. (11 points)Test the hypotheses

versus

, applying

.

b. (11 points) Calculate the power of test if the authentic mean interior temperature is as high since 22. seventy five.

9-54

Terrain: a)

1) The unbekannte of interest is the true indicate interior temperature life, пЃ­. 2) H0: пЃ­ sama dengan 22. your five

3) H1: пЃ­ п‚№ 22. your five

4) t0 пЂЅ

times пЂ­пЃ­

s/ n

5) Reject H0 if |t0| > tпЃЎ/2, n-1 in which пЃЎ sama dengan 0. 05 and tпЃЎ/2, n-1 sama dengan 2 . 776 for and = a few 6) x пЂЅ 22. 496, s = 0. 378, d = your five

t0 пЂЅ

22. 496 пЂ­ 22. 5

zero. 378 as well as 5

пЂЅ пЂ­0. 00237

7) Mainly because –0. 00237 > 2. 776 we cannot reject the null hypothesis. There is not sufficient evidence to conclude that the true suggest interior temperatures is not really equal to twenty two. 5C at  sama dengan 0. 05. b) g =

пЃ¤ | пЃ­ пЂ­ пЃ­0 | | 22. seventy five пЂ­ twenty-two. 5

пЂЅ

пЂЅ

пЂЅ zero. 66

пЃі

пЃі

0. 378

Using the OC contour, Chart VII e) pertaining to пЃЎ sama dengan 0. 05, d sama dengan 0. sixty six, and n = 5, we obtain пЃў пЃЂ zero. 8 and power of 1пЂ­0. 8 sama dengan 0. installment payments on your OR

(

)

в€љ

в€љ

5. (9-82) In the event the standard deviation of pit diameter exceeds 0. 01 millimeters, there is certainly an unacceptably high probability that the rivet will not match. Suppose that and

millimeter.

a. (13 points) Is there solid evidence to indicate that the common deviation of hole size exceeds zero. 01 millimeter? Use

. State any important assumptions

about the underlying distribution of the data.

b. (9 points) Find the P-value just for this test.

Encanto: a) To be able to use the пЃЈ2 statistic in hypothesis screening and self confidence interval structure, we need to imagine the root distribution is normal.

1) The parameter appealing is the the case standard deviation of the diameter, пЃі. However , the answer is available by doing a hypothesis test in пЃі2. 2) H0: пЃі2 = zero. 0001

3) H1: пЃі2 > 0. 0001

2

4) пЃЈ 0 sama dengan

(n пЂ­ 1)s 2

пЃі2

a couple of

2

5) Reject H0 if пЃЈ02 пЂѕ пЃЈпЃЎ, n пЂ­1 where пЃЎ = 0. 01 and пЃЈ0. 01, 14 sama dengan 29. 13 for and = 12-15

6) n = 15, s2 = 0. 008

пЃЈ2

0

=

( n пЂ­ 1)s2

пЃі2

пЂЅ

14(0. 008)2

пЂЅ 8. ninety six

0. 0001

7) Since 8. 96 < twenty nine. 14 do not reject H0. There is insufficient evidence in conclusion that the the case standard change of the diameter exceeds zero. 0001 by пЃЎ sama dengan 0. 01. P-value sama dengan P(пЃЈ2 > 8. 96) for 18 degrees of independence: 0. five < P-value < 0. 9 5.

(11-12) A write-up in the Diary of the American Ceramic Contemporary society [" Rapid Hot-Pressing of Ultrafine PSZ Powders” (1991, Vol. 74, pp. 1547–1553)] considered the microstructure of the ultrafine powder of partially stable zirconia being a function of temperature. The information are displayed below:

a. (15 points) Fit the simple linear regression model using the method of least squares. m. (5 points) Estimate the mean porosity for a temp of 1400 п‚°C. Encanto: a)

The regression formula is

Porosity = 55. 6 -- 0. 0342 Temperature

Predictor

Coef APRENDI Coef

Capital t

P

Constant

55. 63 32. 10 1 . 73 0. a hundred and forty four...



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