Aspecific type of concrete has a normally distributed compressive strength with a mean of 1 psi and standard deviation of 1500 psi. the concrete is used to cast ten beams. the minimum compressive strength expected of each beam is 8750 psi. • determine the probability that two of the ten cylinders will fail the minimum compressive strength requirement? • determine the probability that atleast three cylinders will fail the minimum compressive strength requirement?
Maths term is known as the probabil
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:
[tex]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[/tex]
- 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:
[tex]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[/tex]
- 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:
[tex]ME = Z^*. \frac{sigma}{\sqrt{n} }[/tex]
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.
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a) 3232.1077 < u < 3267.8923
b) 3226.448 <u< 3273.552
Explanation:
[tex]3. a civil engineer is analyzing the compressive strength of concrete. compressive strength is norma[/tex]
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Compressive strength of bricks is the capacity of brick to resist or withstand under compression when tested on Compressive testing machine [CTM]. The Compressive strength of a material is determined by the ability of the material to resist failure in the form of cracks and fissure.
Explanation:
If you want to know more, refer to this link:
https://civilread.com/compressive-strength-test-on-brick/#:~:text=Compressive%20strength%20of%20bricks%20is,form%20of%20cracks%20and%20fissure.
confidence intervals (3232.10 , 3267.89)
Step-by-step explanation:
Explanation:-
Confidence intervals on mean :-
The values χ ± 1.96 б/[tex]\sqrt{n}[/tex] is called the 95% of confidence intervals
Given sample size 'n' = 12
mean value 'x' = 3250
and variance σ² = 1000
now standard deviation σ = [tex]\sqrt{1000} = 31.622[/tex]
now the confidence interval ( χ - 1.96 б/[tex]\sqrt{n}[/tex] , χ + 1.96 б/[tex]\sqrt{n}[/tex] )
substitute given values and simplification , we get
[tex](3250 - 1.96\frac{31.622}{\sqrt{12} } ,3250 +1.96\frac{31.622}{\sqrt{12} })[/tex]
use calculator
95 % of confidence intervals are (3232.10 , (3267.89)
The given mean value is lies between in these confidence intervals
(3232.10 , (3267.89)