Engineering, 23.02.2021 05:10 stephanielynch

. Air entrainment is a process of entrapping tiny air bubbles in concrete mix in order to increase the durability of the hardened concrete in freeze-thaw climates. After 35 days, the breaking stress [in psi] was measured for concrete samples with and without air entrainment. Based on the data, does the air entrainment process increase the breaking stress of the concrete

Answers: 3

Show answers

Answers

Answer from: angelicar4144

This question is incomplete, the complete question is;

Air entrainment is a process of entrapping tiny air bubbles in concrete mix in order to increase the durability of the hardened concrete in freeze-thaw climates. After 35 days, the breaking stress [in psi] was measured for concrete samples with and without air entrainment. Based on the data, does the air entrainment process increase the breaking stress of the concrete. { ∝ = 0.05 }, assuming population variances are equal.

Air Entrainment No Air Entrainment

4479 4118

4436 4531

4358 4315

4724 4237

4414 3888

4358 4279

4487 4311

3984

4197

4327

Since p-value ( 0.088) > significance level ( 0.05)

hence, Failed to reject Null hypothesis

It is then concluded that the null hypothesis H₀ is NOT REJECTED.

Therefore, there is no sufficient evidence to claim that population mean μ1 is greater than μ2 at 0.05 significance level.

We conclude that Air entrainment process can't increase the breaking stress of the concrete.

Explanation:

Given the data in the question;

mean x" = (4479 + 4436 + 4358 + 4724 + 4414 + 4358 + 4487 + 3984 + 4197 + 4327) / 10

mean x"1 = 43764 / 10 = 4376.4

x ( x - x" ) ( x - x" )²

4479 102.6 10526.76

4436 59.6 3552.16

4358 -18.4 338.56

4724 347.6 120825.76

4414 37.6 1413.76

4358 -18.4 338.56

4487 110.6 12232.36

3984 -382.4 153977.76

4197 -179.4 32184.36

4327 -49.4 2440.36

∑ 337830.4

Standard deviation s1 = √( (∑( x - x" )²) / n -1

Standard deviation s1 = √( 337830.4 / (10 - 1 ))

Standard deviation s1 = 193.74

x2 ( x2 - x"2 ) ( x2 - x"2 )²

4118 -121.9 14859.61

4531 291.1 84739.21

4315 75.1 5640.01

4237 -2.9 8.41

3888 -351.9 123833.61

4279 39.1 1528.81

4311 71.1 5055.21

∑ 235664.87

mean x"2 = (4118 + 4531 + 4315 + 4237 + 3888 + 4279 + 4311) / 7

mean x"2 = 29679 / 7 = 4239.9

Standard deviation s2 = √( (∑( x2 - x" )²) / n2 - 1

Standard deviation s1 = √( 337830.4 / (7 - 1 ))

Standard deviation s1 = 198.19

Mean x"1 = 4376.4, S.D1 = 193.74, n1 = 10

Mean x"2 = 4239.9, S.D2 = 198.19, n2 = 7

so;

Null Hypothesis H₀ : μ1 = μ2

Alternative Hypothesis H₁ : μ1 > μ2

Lets determine our rejection region;

based on the data provided. the significance level ∝ = 0.005

with degree of freedom DF = n1 + n2 - 2 = 10 + 7 - 2 = 15

so, Critical Value = 1.753

The rejection region for this right -tailed is R = t:t > 1.753

Test statistics

since it is assumed that the population variances are equal, so we calculate pooled standard deviation;

Sp = √{ [ (n1 -1)S.D1² + (n2 - 1)S.D2²] / [ n1 + n2 -2 ]

we substitute

Sp = √{ [ (10 -1)(193.74)² + (7 - 1)(198.19)²] / [ 10 + 7 -2 ]

Sp = √ [ 573492.345 / 15 ]

Sp = 195.53

so the Test statistics will be;

t = (x"1 - x"2) / Sp√( $\frac{1}{n1}$ + $\frac{1}{n2}$ )

t = (4376.4 - 4239.9) / 195.53√( $\frac{1}{10}$ + $\frac{1}{7}$ )

t = 136.5 / 96.36

t = 1.42

P-value = 0.088

Since p-value ( 0.088) > significance level ( 0.05)

hence, Failed to reject Null hypothesis

It is then concluded that the null hypothesis H₀ is NOT REJECTED.

Therefore, there is no sufficient evidence to claim that population mean μ1 is greater than μ2 at 0.05 significance level.

We conclude that Air entrainment process can't increase the breaking stress of the concrete.

Answer from: Quest

answer:

(b)

explanation:

Answer from: Quest

Answer: i doo!
explanation:

Another question on Engineering

Engineering, 04.07.2019 18:10

Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.4-m-long tubes inside which the refrigerant is evaporating at -20°c. air approaches the tube bank in the normal direction at 0°c and 1 atm with a mean velocity of 4 m/s. the tubes are arranged in-line with longitudinal and transverse pitches of sl- st 1.5 cm. there are 30 rows in the flow direction with 15 tubes in each row. determine (a) the refrigeration capacity of this system and (b) pressure drop across the tube bank. evaluate the air properties at an assumed mean temperature of -5°c and 1 atm. is this a good assumption?

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Adouble-strand no. 60 roller chain is used to transmit power between a 13-tooth driving sprocket rotating at 300 rev/min and a 52-tooth driven sprocket. a) what is the allowable horsepower of this drive? b) estimate the center-to-center distance if the chain length is 82 pitches. c) estimate the torque and bending force on the driving shaft by the chain if the actual horsepower transmitted is 30 percent less than the corrected (allowable) power.

Answers: 3

Answer

Engineering, 04.07.2019 18:10

The filament of an incandescent lamp has a temperature of 2000k. calculate the fraction of radiation emitted in the visible light band if the filament is approximated as blackbody

Answers: 2

Answer

Engineering, 04.07.2019 18:10

Shafts are machine elements that are used to a) carry axial loads b) direct shear loads c) transmit power d) rotate at constant speed e) none of the above circular and square shafts subjected to the same torque under the same circum behave a) the same way b) almost the same way

Answers: 2

Answer

You know the right answer?

. Air entrainment is a process of entrapping tiny air bubbles in concrete mix in order to increase t...

Questions