based on the attached QQ-plot, can I say that the distribution is approximately normal? The results for Shapiro-Wilk test is p-value = 0.1334 but the sample size is small (only 12 as they are observations for each month) so I am not sure if its suitable to use the results from Shapiro test.
No, it is not. As the line has a different slope by the line of y = x, it means that this distribution has a different skewness than the normal distribution.
Related
Random sample of 143 girl and 127 boys were selected from a large population.A measurement was taken of the haemoglobin level(measured in g/dl) of each child with the following result.
girl n=143 mean = 11.35 sd = 1.41
boys n=127 mean 11.01 sd =1.32
estimate the standard error of the difference between the sample means
In essence, we'd pool the standard errors by adding them. This implies that we´re answering the question: what is the vairation of the sampling distribution considering both samples?
SD = sqrt( (sd₁**2 / n₁) + (sd₂**2 / n₂) \
SD = sqrt( (1.41**2 / 143) + (1.32**2 / 127) ≈ 0.1662
Notice that the standrad deviation squared is simply the variance of each sample. As you can see, in our case the value is quite small, which indicates that the difference between sampled means doesn´t need to be that large for there to be a larger than expected difference between obervations.
We´d calculate the difference between means as 0.34 (or -0.34 depending on the nature of the question) and divide this difference by the standrad error to get a t-value. In our case 2.046 (or -2.046) indicates that the observed difference is 2.046 times larger than the average difference we would expect given the variation the variation that we measured AND the size of our sample.
However, we need to verify whether this observation is statistically significant by determining the t-critical value. This t-critical can be easily calculated by using a t-value chart: one needs to know the alpha (typically 0.05 unless otherwise stated), one needs to know the original alternative hypothesis (if it was something along the lines of there is a difference between genders then we would apply a two tailed distribution - if it was something along the lines of gender X has a hameglobin level larger/smaller than gender X then we would use a single tailed distribution).
If the t-value > t-critical then we would claim that the difference between means is statistically significant, thereby having sufficient evident to reject the null hypothesis. Alternatively, if t-value < t-critical, we would not have statistically significant evidence against the null hypothesis, thus we would fail to reject the null hypothesis.
I would like to know if there is any standard algorithm or statistical parameter than can be used to determine how many minimum samples should be considered from beginning whose average value nearly matches with the average of all samples.
For Example : If 2000 samples are present and Average is 20
Acceptable average range is 20+-0.01
If we start taking average from first sample then by taking average of X samples we can get average within 20+-0.01
Problem is about find value of X
Just need guidance from logical perspective [Procedure or Algorithm to consider]
Thanks in advance
Alright, so if the standard deviation is known, then for a 95% confidence that the sample mean will be within 0.01 of the true mean for a normal distribution with standard deviation equal to s, we require that:
0.01 = z95 x s / sqrt(n)
Here, z95 is the two-sided CDF of the normal distribution and is about 1.96 (from tables), s is the standard deviation and n is the number of samples required. We can solve for n in terms of s:
0.01 = 1.96 x s / sqrt(n)
<=> sqrt(n) = 196s
<=> n = 38416s
So, if s = 1, you'd expect to need about 38.5k samples to get a 95% confidence that the sample mean would be within 0.01 of the true mean. The number of samples needed to achieve a given precision is directly proportional to the true sample standard deviation.
If the true population's standard deviation is not known, the calculation works in a similar fashion, except you would use the CDF from the student's T distribution (so instead of z95 you'd use t95) and you'd use the sample standard deviation.
If you want a different confidence interval - higher or lower - you'd look up the corresponding two-sided CDF for whichever distribution you're using and use the corresponding value (so something besides 1.96).
The discussion on Wikipedia, Basic Steps section, is instructive: https://en.wikipedia.org/wiki/Confidence_interval
I'm writing a program that lets users run simulates on a subset of data, and as part of this process, the program allows a user to specify what sample size they want based on confidence level and confidence interval. Assuming a p value of .5 to maximum sample size, and given that I know the population size, I can calculate the sample size. For example, if I have:
Population = 54213
Confidence Level = .95
Confidence Interval = 8
I get Sample Size 150. I use the formula outlined here:
https://www.surveysystem.com/sample-size-formula.htm
What I have been asked to do is reverse the process, so that confidence interval is calculated using a given sample size and confidence level (and I know the population). I'm having a horrible time trying to reverse this equation and was wondering if there is a formula. More importantly, does this seem like an intelligent thing to do? Because this seems like a weird request to me.
I should mention (just to be clear) that the CI is estimated for the mean, not the population. In that case, if we assume the population is normally distributed and that we know the population standard deviation SD, then the CI is estimated as
From this formula you would also get your formula, where you are estimating n.
If the population SD is not known then you need to replace the z-value with a t-value.
I am stuck in a statistics assignment, and would really appreciate some qualified help.
We have been given a data set and are then asked to find the 10% with the lowest rate of profit, in order to decide what Profit rate is the maximum in order to be considered for a program.
the data has:
Mean = 3,61
St. dev. = 8,38
I am thinking that i need to find the 10th percentile, and if i run the percentile function in excel it returns -4,71.
However I tried to run the numbers by hand using the z-score.
where z = -1,28
z=(x-μ)/σ
Solving for x
x= μ + z σ
x=3,61+(-1,28*8,38)=-7,116
My question is which of the two methods is the right one? if any at all.
I am thoroughly confused at this point, hope someone has the time to help.
Thank you
This is the assignment btw:
"The Danish government introduces a program for economic growth and will
help the 10 percent of the rms with the lowest rate of pro t. What rate
of pro t is the maximum in order to be considered for the program given
the mean and standard deviation found above and assuming that the data
is normally distributed?"
The excel formula is giving the actual, empirical 10th percentile value of your sample
If the data you have includes all possible instances of whatever you’re trying to measure, then go ahead and use that.
If you’re sampling from a population and your sample size is small, use a t distribution or increase your sample size. If your sample size is healthy and your data are normally distributed, use z scores.
Short story is the different outcomes suggest the data you’ve supplied are not normally distributed.
I've got files with irradiance data measured every minute 24 hours a day.
So if there is a day without any clouds on the sky the data shows a nice continuous bell curves.
When looking for a day without any clouds in the data I always plotted month after month with gnuplot and checked for nice bell curves.
I was wondering If there's a python way to check, if the Irradiance measurements form a continuos bell curve.
Don't know if the question is too vague but I'm simply looking for some ideas on that quest :-)
For a normal distribution, there are normality tests.
In short, we abuse some knowledge we have of what normal distributions look like to identify them.
The kurtosis of any normal distribution is 3. Compute the kurtosis of your data and it should be close to 3.
The skewness of a normal distribution is zero, so your data should have a skewness close to zero
More generally, you could compute a reference distribution and use a Bregman Divergence, to assess the difference (divergence) between the distributions. bin your data, create a histogram, and start with Jensen-Shannon divergence.
With the divergence approach, you can compare to an arbitrary distribution. You might record a thousand sunny days and check if the divergence between the sunny day and your measured day is below some threshold.
Just to complement the given answer with a code example: one can use a Kolmogorov-Smirnov test to obtain a measure for the "distance" between two distributions. SciPy offers a neat interface for this, called kstest:
from scipy import stats
import numpy as np
data = np.random.normal(size=100) # Our (synthetic) dataset
D, p = stats.kstest(data, "norm") # Perform a one-sided Kolmogorov-Smirnov test
In the above example, D denotes the distance between our data and a Gaussian normal (norm) distribution (smaller is better), and p denotes the corresponding p-value. Other distributions can be similarly tested by substituting norm with those implemented in scipy.stats.