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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 75 kg bungee jumper leaps from a bridge. When he is 30 meters above the water, and moving at a speed of 20 m/s, the bungee cor
jasenka [17]

Answer:

k = 52.2 N / m

Explanation:

For this exercise we are going to use the conservation of mechanical energy.

Starting point. When it is 30 m high

        Em₀ = K + U = ½ m v² + m g h

Final point. Right when you hit the water

        Em_{f} = K_{e} = ½ k x²

in this case the distance the bungee is stretched is 30 m

        x = h

as they indicate that there are no losses, energy is conserved

        Em₀ = Em_{f}

       ½ m v² + m g h = ½ k h²

       k = \frac{m (v^{2} + 2 g h)}{h^{2} }

let's calculate

       k = \frac{75 \ ( 20^{2}  + 2 \ 9.8 \ 30)}{30^{2} }

       k = 52.2 N / m

3 0
3 years ago
until a train is a safe distance from the station, it must travel at 5 m/s. once the train is on open track, it can speed up to
MatroZZZ [7]

Answer:

5 meters per second squared

Explanation:

We calculate the acceleration using the formula:

a = (vf - vi) / t

where "vf" is the final velocity, "vi" the initial velocity, and "t" the time it took to change from the initial velocity to the final one.

In our case:

a = (45 - 5) / 8 = 40 / 8 = 5 m/s^2

4 0
4 years ago
Which does a reference point provide
mr_godi [17]
Point of reference - an indicator that orients you generally; "it is used as a reference for comparing the heating and the electrical energy involved"
5 0
3 years ago
Need help with these two questions Please?
Ksivusya [100]
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5 0
3 years ago
A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial
V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF=2.1\times 10^{-12} F

1pF=10^{-12} F

d=5.0 mm=5\times 10^{-3} m

1mm=10^{-3} m

a.Maximum flux

\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}

\epsilon_0=8.85\times 10^{-12}

\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm

b.Maximum displacement current,I=\frac{V}{R}=\frac{30}{130}=0.23 A

c.We have to find electric flux at t=0.5 ns

t=0.5ns=0.5\times 10^{-9} s

1ns=10^{-9}s

q=CV(1-e^{-\frac{t}{RC}})

q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})

q=52.9\times 10^{-12} C

\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm

d.Displacement current at t=0.5ns

I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}

I=0.037 A

5 0
3 years ago
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