A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a true average compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. State the relevant hypotheses, and describe the type I and type II errors in the context of this problem.

Answers

Answer 1

Answer:

Null Hypothesis - True average compressive strength of  mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.

Type I error (false positive) - this occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.

Type II error (false-negative)- his occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is  more than 1300 KN/m2 in actual scenario but the measurements show an incorrect  reading of True average compressive strength less than 1300 KN/m2.

Explanation:

The hypothesis for this piece of information is as follows -

Null Hypothesis - True average compressive strength of  mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.

Alternate hypothesis - True average compressive strength of  mixture of pulverized fuel ash and Portland cement is less than or equal to 1300 KN/m2

Type I error (false positive) - this occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.

Type II error (false-negative)- his occur when the True average compressive strength of  mixture of pulverized fuel ash and Portland cement is  more than 1300 KN/m2 in actual scenario but the measurements show an incorrect  reading of True average compressive strength less than 1300 KN/m2.


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A resistor, inductor, and capacitor are in parallel in a circuit where the frequency of operation can vary. The R, L, and C values are such that at the frequency omega subscript 0, the magnitude of all the impedances are equal to each other. If the frequency of operation approaches zero, which element will dominate in determining the equivalent impedance of this parallel combination?

a. The inductor.
b. The capacitor.
c. The resistor.
d. Insufficient information provided.

Answers

Answer:

Option A is correct

Explanation:

As we know  

Inductive Susceptance = ½(pi)*f*L

Or  Inductive Susceptance is inversely  proportional to the frequency

Likewise conductive Susceptance = 2 (pi)*f*C

Conductive Susceptance is directly proportional to the frequency

When the frequency will reach the value zero, then the Inductive Susceptance will become infinite

Hence, inductor will dominate in determining the equivalent impedance of this parallel combination

Option A

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