1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
klio [65]
3 years ago
7

A civil engineer is analyzing the compressive strength of concrete. The compressive strength is approximately normal distributed

with variance 1000 psi2. A random sample of 12 specimens has a compressive strength of 3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237, 3286, 3210, 3265, and 3273 psi. (a) Test the hypothesis that the average compressive strength is 3500 psi. Use a fixed level test with 0.01. (b) What is the smallest level of significance at which you would be willing to reject the null hypothesis? (c) Construct a 95% two-sided IC at medium-force compression. (d) Construct a 99% two-sided IC at medium-force compression. Compare the width of this confidence interval with the width of that found in part
Engineering
1 answer:
hram777 [196]3 years ago
7 0

Answer:

See explanation

Explanation:

Solution:-

- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).

- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.

- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:

            3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,

                          3286, 3210, 3265, 3273

- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.

      Null hypothesis: μ = 3500 psi

      Alternate hypothesis: μ ≠ 3500 psi

- The type of test performed on the sample data will depend on the application of Central Limit Theorem.

- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )

- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.

- The mean of the sample ( x ) is calculated as follows:

    x = \frac{Sum ( x_i )}{n} \\\\x = \frac{Sum ( 3273+ 3229+ 3256+ 3272+ 3201+ 3247+ 3267+ 3237+ 3286+ 3210+ 3265+3273 )}{12} \\\\x = \frac{39016}{12} \\\\x = 3251.3333

 

- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.

- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:

  • The population variance is known OR sample size ( n ≥ 30 )    

- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:

              Z-test = \frac{x - u}{\sqrt{\frac{sigma^2}{n} } } \\\\Z-test = \frac{3251.333 - 3500}{\sqrt{\frac{1000^2}{12} } } \\\\Z-test = -27.24      

- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:

           Z-critical = Z_α/2 = Z_0.005

- Use the list of correlation of significance level ( α ) and critical values of Z to determine:

          Z-critical = Z_0.005 = ± 2.576

- Compare the Z-test value against the rejection region defined by the Z-critical value.

     Rejection region: Z > 2.576 or Z < -2.576

- The Z-test value lies in the rejection region:

            Z-test < Z-critical

           -27.24 < -2.576 .... Null hypothesis rejected

Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.

- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.

- The margin of error (ME) is defined by the following formula:

              ME = Z^*. \frac{sigma}{\sqrt{n} }

Where,

- The ( Z* ) is the critical value for the defined confidence level ( CI ):

- The confidence interval and significance level are related and critical value Z* is as such:

   

            α = 1 - CI , Z* = Z_α/2

- The critical values for ( CI = 99% & 95% ) are evaluated:

           α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05

           Z* = Z_0.005        ,   Z* = Z_0.025

           Z* = ± 2.58            ,   Z* = ± 1.96

- The formulation of Confidence interval is given by the following inequality:

                 [ x - ME  <    μ    <   x + ME ]

                 [ x - Z*√σ^2 / n   <    μ    <   x + Z*√σ^2 / n ]

- The CI of 95% yields:

   [ 3251.33 - 1.96*√(1000 / 12)   <    μ    <   3251.33 + 1.96*√(1000 / 12) ]

                [ 3251.333 - 17.89227 <    μ    <   3251.33 + 17.89227 ]

                              [ 3233.44  <    μ    <  3269.23  ]

- The CI of 99% yields:

   [ 3251.33 - 2.58*√(1000 / 12)   <    μ    <   3251.33 + 2.58*√(1000 / 12) ]

                [ 3251.333 - 23.552 <    μ    <   3251.33 + 23.552 ]

                              [ 3227.78  <    μ    <  3274.88  ]

                 

- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.    

You might be interested in
Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...
leva [86]

Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

Explanation:

Is the same as the example,

If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

Then

(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)

Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...

The way to write this is

Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)

(photo)

6 0
3 years ago
What are the two main what are the two main concerns in the research of fluid power efficiency?
Galina-37 [17]

Answer:

The correct option is;

Materials and Components

Explanation:

The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on

1) Advances in fluid power

2) Making use of the advantages

3) Making use of the other externally available technological advantages

4) Giving allowance for disadvantages

Areas of interest in advances in fluid power are;

a. Computer optimized flow

b. The use of new and improved materials/coatings

c. The use of components that save energy, such as intelligent supply pressure adapting systems

3 0
2 years ago
Read 2 more answers
The period of an 800 hertz sine wave is
sukhopar [10]

Explanation:

White Board Activity: Practice: A sound has a frequency of 800 Hz. What is the period of the wave? The wave repeats 800 times in 1 second and the period of the function is 1/800 or 0.00125.

3 0
2 years ago
Select the correct answer. Which existing technology did engineers use to enhance the speed of propeller-driven airplanes
Musya8 [376]

metallurgy:

Explanation:

7 0
2 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
Other questions:
  • I need solution for this question please ​
    7·2 answers
  • What is the capacity of the machine in batches?
    10·1 answer
  • What kind or kinds of engineers does take to design a drone and why?
    11·1 answer
  • Why does an aeroplane smoke in the air​
    14·1 answer
  • What are the important things to remember when arriving for an interview?
    15·1 answer
  • A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut
    14·1 answer
  • Harlin is designing a new car engine that does not create pollution. Which technological design factor is probably the most
    8·1 answer
  • Bob would like to run his house off the grid, therefore he needs to find out how many solar panels and batteries he needs to buy
    12·1 answer
  • Which - type of service shop is least likely to provide service to all
    9·1 answer
  • Which of following is not malicious ?<br> Worm<br> Trogan Horse<br> Driver<br> Virus
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!