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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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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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