Yes. The standard deviation and mean would be less. How much less would depend on the sample size, the distribution that the sample was taken from (parent distribution) and the parameters of the parent distribution. The affect on the sampling distribution of the mean and standard deviation could easily be identified by Monte Carlo simulation.
Yes
Intuitively, a standard deviation is a change from the expected value.For the question you asked, this means that the change in the "results" doesn't exist, which doesn't really happen. If the standard deviation is 0, then it's impossible to perform the test! This shows that it's impossible to compute the probability with the "null" standard deviation from this form:z = (x - µ)/σIf σ = 0, then the probability doesn't exist.
Not necessarily. The standard deviation measures (in simplified terms) how different the numbers are from each other, while the mean is their average. If the standard deviation decreases, it means the numbers are closer to each other, it doesn't change how big the numbers are.
This would increase the mean by 6 points but would not change the standard deviation.
The increase in sample size will reduce the confidence interval. The increase in standard deviation will increase the confidence interval. The confidence interval is not based on a linear function so the overall effect will require some calculations based on the levels before and after these changes. It would depend on the relative rates at which the change in sample size and change in standard deviation occurred. If the sample size increased more quickly than then standard deviation, in some sense, then the size of the confidence interval would decrease. Conversely, if the standard deviation increased more quickly than the sample size, in some sense, then the size of the confidence interval would increase.
Yes
They will differ from one sample to another.
Intuitively, a standard deviation is a change from the expected value.For the question you asked, this means that the change in the "results" doesn't exist, which doesn't really happen. If the standard deviation is 0, then it's impossible to perform the test! This shows that it's impossible to compute the probability with the "null" standard deviation from this form:z = (x - µ)/σIf σ = 0, then the probability doesn't exist.
Not necessarily. The standard deviation measures (in simplified terms) how different the numbers are from each other, while the mean is their average. If the standard deviation decreases, it means the numbers are closer to each other, it doesn't change how big the numbers are.
This would increase the mean by 6 points but would not change the standard deviation.
They would both increase.
T-scores have a mean of 50 and a standard deviation of 10. These values are fixed and do not change regardless of the distribution of T-scores.
The increase in sample size will reduce the confidence interval. The increase in standard deviation will increase the confidence interval. The confidence interval is not based on a linear function so the overall effect will require some calculations based on the levels before and after these changes. It would depend on the relative rates at which the change in sample size and change in standard deviation occurred. If the sample size increased more quickly than then standard deviation, in some sense, then the size of the confidence interval would decrease. Conversely, if the standard deviation increased more quickly than the sample size, in some sense, then the size of the confidence interval would increase.
If I take 10 items (a small sample) from a population and calculate the standard deviation, then I take 100 items (larger sample), and calculate the standard deviation, how will my statistics change? The smaller sample could have a higher, lower or about equal the standard deviation of the larger sample. It's also possible that the smaller sample could be, by chance, closer to the standard deviation of the population. However, A properly taken larger sample will, in general, be a more reliable estimate of the standard deviation of the population than a smaller one. There are mathematical equations to show this, that in the long run, larger samples provide better estimates. This is generally but not always true. If your population is changing as you are collecting data, then a very large sample may not be representative as it takes time to collect.
Consider thatxd= x- Arithmetic meand211-1.5 = -0.50.2522-1.5 = 0.50.25=0.5Arithmetic mean = (1+2)/2 =1.5Standard deviation=ie .5= 0.70now Considerxd= x- Arithmetic meand211-2=-1122-2=0033-2=11Arithmetic mean= (1+2+3)/3 = 2 =2Standard deviation= = (2/2) = 1So the Standard deviation can increasenow Considerxd= x- Arithmetic meand211-1.25=-0.250.062522-1.25=0.750.562511-1.25=-0.250.062511-1.25=-0.250.0625Arithmetic mean= (1+2+1+1)/4= 1.25 = .75Standard deviation= = (0.75/4) = 0.4330So the Standard deviation can decreaseStandard deviation can either decrese or increase or remains the same
Click on the deviation you no longer want favorited, then once you are on deviation's page you look at the area where it would normally say "Add to favorites". If it was already favorited by you it will then say "Remove from favorites". Click that and wait until it says "Removed" . If you change your mind and want to favorite it again - just wait and the "Removed" button will go back to "Add to favorites".
Click on the deviation you no longer want favorited, then once you are on deviation's page you look at the area where it would normally say "Add to favorites". If it was already favorited by you it will then say "Remove from favorites". Click that and wait until it says "Removed" . If you change your mind and want to favorite it again - just wait and the "Removed" button will go back to "Add to favorites".