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Andreyy89
1 year ago
7

What is the average value expected for the independent-measures t statistic if the null hypothesis is true?

Physics
1 answer:
Bogdan [553]1 year ago
5 0

The average value for t- statistics is 0 if null hypothesis is true.

What is a t – statistics and null hypothesis ?

  • The t-statistic, which is used in statistics, measures how far a parameter's estimated value deviates from its hypothesized value relative to its standard error.
  • Through the Student's t-test, it is utilized in hypothesis testing. In a t-test, the t-statistic is used to decide whether to accept or reject the null hypothesis.
  • It is quite comparable to the z-score, but when the data is small or the standard deviation is unknown, the t-statistic is employed instead.
  • The default assumption, or null hypothesis, is that the quantity being measured is zero (null). The quantity to be assessed is often the difference between two circumstances.
  • Trying to determine, for instance, whether there is conclusive evidence that an impact has happened or that samples come from several batches

The t-test yields a t-value of 0 if the sample data precisely match the null hypothesis. The value of the t-value rises as the sample data diverge more and more from the null hypothesis.

To learn more about null hypothesis visit here:

brainly.com/question/25263462

#SPJ4

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A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
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Answer:

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Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

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V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

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