My data set has an age range variable, but I would like to calculate the mean and standard deviation of age.
Since your data is categorical, there isn't a way to calculate the "true" sample mean and standard deviation of respondent age. There are a few different ways you could estimate, depending on how sophisticated you'd like to get.
The simplest way would be to assign an age to each band (say, the mid-point) and summarize on that. The downside is that you will be underestimating the standard deviation (clumping data together tends to do that). To the extent your categories are not uniformly distributed (and from your image they don't appear to be), your estimate of the mean will also be off.
* set point estimates for each age band .
RECODE age (1=22) (2=30) (3=40) (4=50) (5=60) (6=70) (7=80) .
EXE .
* calculate mean and std dev .
MEANS age /CELLS MEAN STDDEV .
More sophisticated estimation techniques might try to account for skews in data (e.g. your sample seems to skew younger) and convert each age band into its own distribution.
For example, instead of assuming 203 respondents are age 22 (as is done in the code above), you might assume 25 respondents each are 18, 19, 20, ... 25. More realistically than that even, you might assume that even that distribution skews younger (e.g. 50 18-yr olds, 40 19-yr old, etc etc).
Automated approaches to that would be interesting as its own question. :)
Related
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 have many measurements of age of the same person. Let's say:
[23 25 32 23 25]
I would like to output a single value and a reliability score of this value. The single value can be the average.
Reliability, I don't know well how to calculate it. The value should be between 0 and 1, where 1 means all ages are equal and a very unreliable measurement should be near 0.
Probably the variance should be used here, but it's not clear to me how to normalize it between 0 and 1 in a meaningful way (1/(x+1) is not much meaningful :)).
Assume some probability distribution (or determine what probability distribution your data fits most accurately). A good choice is a normal distribution, which for discrete data requires a continuity correction. See example here: http://www.milefoot.com/math/stat/pdfc-normaldisc.htm
In your example, your reliability score for the average age of 26 (25.6 rounded to nearest integer), is simply the probability that X falls in the range (25.5, 26.5).
The easiest way for assessing reliability (or internal consistency) is to use Cronbach's alpha. I guess most statistics software has this method built-in.
https://en.wikipedia.org/wiki/Cronbach%27s_alpha
I have a requirement where I have to verify the transmit power out of a device as measured at its connector is within 2 dB of its expected value over 95% of test measurements.
I am using a signal analyzer to analyze the transmitted power. I only get the average power value, min, max and stdDev of the measurements and not the individual power measurements.
Now, the question is how would I verify the "95% thing" using average power, min, max and stdDev. It seems that I can use normal distribution to find the 95% confidence level.
I would appreciate if someone can help me on this.
Thanks in anticipation
The way I'm reading this, it seems you are a statistical beginner, so if I'm wrong there, the rest of this answer will probably be insultingly basic, and I'm sorry.
Anyway, the idea is that if a dataset is normally distributed, and all the observations are independent of one another, then 95% of the data points will fall within 1.96 standard deviations of the mean.
Do you get identical estimates of average power every time you measure, or are there some slight random differences from reading to reading? My guess is that it's the second. If you were to measure the power a whole bunch of times, and each time you plotted your average power value on a histogram, then that histogram of sample means would have the shape of a bell curve. This bell curve of sample means would have its own mean and standard deviation, and if you have thousands or millions of data points going into the calculation of each average power reading, it's not horrible to assume that it is a normal distribution. The explanation for this phenomenon is known as the 'central limit theorem', and I recommend both the Khan academy's presentation of it as well as the wikipedia page on it.
On the other hand, if your average power is the mean of some small number of data points, like for instance n= 5, or n= 30, then assumption of a normal distribution of sample means can be pretty bad. In this case, your 95% confidence interval around the average power goes from qt(0.975,n-1)*SD/sqrt(n) below the average to qt(0.975,n-1)*SD/sqrt(N) above the average, where qt(0.975,n-1) is the 97.5th percentile of the t distribution with n-1 degrees of freedom, and SD is your measured standard deviation.
I'm trying to find confidence intervals for the means of various variables in a database using SPSS, and I've run into a spot of trouble.
The data is weighted, because each of the people who was surveyed represents a different portion of the overall population. For example, one young man in our sample might represent 28000 young men in the general population. The problem is that SPSS seems to think that the young man's database entries each represent 28000 measurements when they actually just represent one, and this makes SPSS think we have much more data than we actually do. As a result SPSS is giving very very low standard error estimates and very very narrow confidence intervals.
I've tried fixing this by dividing every weight value by the mean weight. This gives plausible figures and an average weight of 1, but I'm not sure the resulting numbers are actually correct.
Is my approach sound? If not, what should I try?
I've been using the Explore command to find mean and standard error (among other things), in case it matters.
You do need to scale weights to the actual sample size, but only the procedures in the Complex Samples option are designed to account for sampling weights properly. The regular weight variable in Statistics is treated as a frequency weight.