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dimulka [17.4K]
3 years ago
5

A small difference between means may not be statistically significant, but it could reach statistical significance with a large

sample becausea) As N increases, the standard error gets smaller, reflecting less variability in sample means, which allows greater sensitivity to detect small but significant differences.b) As N increases, distributions get more heterogeneous, allowing small differences to be detected.c) The overlap between distributions grows as sample size increases.d) The difference between means gets larger as sample size increases
Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer:

Increasing the sample size could make a small difference between means reach statistical significance because;

a) As N increases the standard error gets smaller reflecting less variability in sample means, which allows greater sensitivity to detect small but significant differences

Explanation:

The p-value of a test statistic is used to express the level of significance of the test result. A significant result is typically considered as one with a small p-value of less than or equal to 0.05 at which point we reject the null hypothesis

The p-value can be made smaller by either

1) Increasing the size of the sample (N)

2) Decreasing the standard error

3) Increasing the difference between the hypothesized parameter and the statistic of the sample

The standard error, SE, is given as follows;

SE = \dfrac{\sigma}{\sqrt{N} }

Where;

σ = The standard deviation of the sample

n = The sample size

Therefore the standard error decreases, gets smaller,  with increasing sample size, N

Therefore, the small difference between the sample mean could reach statistical significance with a large sample because as N increases the standard error gets smaller reflecting less variability in sample means, which allows greater sensitivity to detect small but significant differences.

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Answer:

- Decreasing the resistance

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Resistance here has a formula of;

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8 0
3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

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Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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Answer:

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ΔФ = BACosθ

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A is the area of the loop

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