In the context of sampling, Bessel’s Correction improves the estimate of standard deviation: specifically, while the sample standard deviation is a biased estimate of the population standard deviation, the bias is smaller with Bessel’s Correction.
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For data, we will use the diamonds data set in the R-Package ggplot2, which contains data from 53940 round cut diamonds. Here are the first 6 rows of this data set:
## # A tibble: 6 x 10
## carat cut color clarity depth table price x y z
## <dbl> <ord> <ord> <ord> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
## 1 0.23 Ideal E SI2 61.5 55 326 3.95 3.98 2.43
## 2 0.21 Premium E SI1 59.8 61 326 3.89 3.84 2.31
## 3 0.23 Good E VS1 56.9 65 327 4.05 4.07 2.31
## 4 0.29 Premium I VS2 62.4 58 334 4.20 4.23 2.63
## 5 0.31 Good J SI2 63.3 58 335 4.34 4.35 2.75
## 6 0.24 Very Good J VVS2 62.8 57 336 3.94 3.96 2.48
Answer this question: what is the meaning of a distribution of a variable, and how does it relate to price?
The distribution of a variable is a description of the relative numbers of times each possilbe outcomes will occur in a number of trials. It relates to price because as it increased the number of times it will occur will decrease.
Explain what a quantitative variable is, and why it was important to make such a choice in a report about standard deviation. Explain how the concepts of numerical and quantitative variables are different, though related.
A quantitative variable is a variable that is measured on a numeric scale. It is important to make such a report about standard deviation because it is important to recognize the outliers in a set of data. It is a number that describes data. Numerical variables are numbers, they can be further classified into discrete and continuous variables. While, quantitative data is expressed using a certain quantity, amount or range.These two are related due to the fact that they both use numbers and describe quantities in data.
What is a histogram? Explain graph below.
A histogram is diagram of rectangles whose area is proportional to the frequency of a variable and whose width is equal to the class interval. This graph is scewed to the right, as the price of diamonds increases the count of diamonds decreases.
Explain the relationship between a histogram and a violin plot.
The violin plot is a method of plotting numeric data, they show the probability density of the data at different values - they usually include the marker for the median of the data indicating the interquartile range. They represent a comparison of variable distribution across different categories.Shows distribution (just like histogram) buy its a histogram on its side, smoothed out, and mirrored on the other side. A histogram is a display of data that uses rectangles to show the frequency of data in successive numerical intervals of equal size.The independant varialbe is plotted along the horizontal axis, while the dependant variable is plotted along the vertical axis. Describes each of the 5 summaries.
R has a function that returns numerical summaries of data. For example:
summary(diamonds$price)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 326 950 2401 3933 5324 18823
Describe what each of these numbers means.
These numbers are the min, 1st Qu., median, mean, 3rd Qu., and max of the price of diamonds. The min is Q0 - the lowest # on the data point. 1st Qu. - is the median of the first half of data. median is the average of all the data points. 3rd Qu. - the median of the second half of data. and the max is the last data point on the scale (the highest data point).
Describe the relationship of the numbers above to the modified box plot, here drawn inside the violin plot. Explain the difference between a boxplot and a modified box plot. Explain what an outlier is, and how suspected outliers are identified in a modified box plot.
The numbers above are shown as the minimum price for diamonds, the 1st Qu., the median, mean, 3rd Qu., and max of the price for diamonds. The graph is showing that diamonds at lower prices are more common then diamonds at higher prices and the box plot includes the outliers, min and max. A modified boxplot is a data display that shows the 5-number summary, its used to find suspected outliers. The whiskers stretch outward away from the 1st Qu. and 3rd Qu. A box plot inlcudes the min, 1st Qu., median, 3rd Qu., and max. An outlier is an observation point that is distant from the other observations.
Add one sentence to indicate where the mean is on this plot.
the red dot indicates the mean
Explain the formulas below, say which uses Bessel’s correction.
The formula used to find the standard deviation. The first one is Bessels correction - (n-1). This is better used, for accuracy, in smaller populations.
\[s = \sqrt{\frac{1}{n-1}\sum\left(x_i - \bar x\right)^2}\]
\[s_n = \sqrt{\frac{1}{n}\sum\left(x_i - \bar x\right)^2}\]
We compute the standard deviation (with Bessel’s correction) of the price variable:
sd(diamonds$price)
## [1] 3989.44
How about without Bessel’s correction? Well, R doesn’t seem to have this function, but we can add it:
sdn <- function(x) {
return(sqrt(mean((x - mean(x))^2)))
}
sdn(diamonds$price)
## [1] 3989.403
How close are these estimates? Which is larger?
these estimates are very close, 1 is slightly larger.
So what is the big deal about Bessel’s correction? See below.
The statement that began this document asserted that Bessel’s correction is important in the context of sampling. Explain sampling here: explain the differences between a population, and a sample, and between a parameter and a statistic. Give examples of parameters and give examples of statistics. Explain the difference between the sample mean and the population mean. Explain the difference between the sample standard deviation and the population standard deviation.
A parameter is a numerical measure that describes a population.A statistic is a numerical value that describes a sample. A sample is collected data from a population. A population refers to the total set of observations that can be made. Example of a parameter: Only 10% of US senators voted for a certain topic. (only 100 US Senators). Example of a statistic: 60% of US citizens agree with the new health policy. (Us citizens are a large populaiton). Sample stardard deviation is a stat that measures the dispersion of data around sample mean. The population sd is the square root of the variance.
We can sample from the diamonds data set and display the price of the diamonds in the sample.
First, we need to choose a sample size, \(n\). We choose \(n=4\) which is very low in practice, but will serve to make a point.
sample.size <- 4
Sampling is random, so next we set the seed. Explain what a seed of a random number generator is. Explain what happens when you use the same seed and what happens when you use different seeds. The simulations below may help you.
Seed is a label for the starting point on the list of sequence of numbers. If you use the same seed youll get the same mean, but if you use different seeds youll get different means.
set.seed(1)
Now let’s try sampling, once.
sample(diamonds$price, sample.size)
## [1] 5801 8549 744 538
Explain what this command did.
gave the sample mean.
Let’s try it with another seed:
set.seed(2)
sample(diamonds$price, sample.size)
## [1] 4702 1006 745 4516
And another:
set.seed(3)
sample(diamonds$price, sample.size)
## [1] 4516 1429 9002 7127
And back to the first one:
set.seed(1)
sample(diamonds$price, sample.size)
## [1] 5801 8549 744 538
Explain these results.
Finally, what happens when we don’t set a seed, between samples.
The sample repeats.
set.seed(1)
sample(diamonds$price, sample.size)
## [1] 5801 8549 744 538
sample(diamonds$price, sample.size)
## [1] 4879 1976 2322 907
sample(diamonds$price, sample.size)
## [1] 463 3376 4932 4616
set.seed(1)
sample(diamonds$price, sample.size)
## [1] 5801 8549 744 538
sample(diamonds$price, sample.size)
## [1] 4879 1976 2322 907
sample(diamonds$price, sample.size)
## [1] 463 3376 4932 4616
Explain these results.
set a seed for 1 and then take the mean for the sample and then the mean of that result.
set.seed(1)
mean(sample(diamonds$price,sample.size))
## [1] 3908
mean(sample(diamonds$price,sample.size))
## [1] 2521
mean(sample(diamonds$price,sample.size))
## [1] 3346.75
Explain what we have done here. Answer the following question: what other statistics could we use to describe samples?
set the seed of 1 then took the standard deviation of that result and then took the standard deviation again from that result.
For example standard deviation, with Bessel’s correction:
set.seed(1)
sd(sample(diamonds$price,sample.size))
## [1] 3936.586
sd(sample(diamonds$price,sample.size))
## [1] 1683.428
sd(sample(diamonds$price,sample.size))
## [1] 2036.409
And standard deviation, without Bessel’s correction:
set.seed(1)
sdn(sample(diamonds$price,sample.size))
## [1] 3409.183
sdn(sample(diamonds$price,sample.size))
## [1] 1457.891
sdn(sample(diamonds$price,sample.size))
## [1] 1763.582
Explain what a sampling distribution of a statistic is and how it relates to the numbers computed above. Answer the following question: what tools do we have to describe these distributions?
It is the distribution of a statistic given in a random sample. The seeds are random numbers so this corresponds to the random sample.
The plot below shows images of the sampling distribution for the sample mean, for different values of sample size.
Answer the following questions: what do the features of the graph below represent? One hint: the horizontal line is the population mean of the prices of all diamonds in the data set.
The line is the population mean.
Explain the concept of an estimator. What is the sample mean estimating, and it what situation does it do a better job?
an estimator = sampling a set of data. A sample mean is an estimate of the population mean.
Let’s try describing the sampling distribution of the sample standard deviation with Bessel’s Correction. Again the samples are of diamonds, and the variable considered is the price of diamonds:
Some people argue that it is appropriate to drop Bessel’s correction for populations, but if the population size is large, as shown here it doesn’t matter much. Why? What is the sample standard deviation estimating? In what situations is it a better estimate?
it is estimating the standard deviation for each sample size. The best estimate occurs at the 1024 sample size.
Now let’s try without Bessel’s correction:
Answer the following questions: what is the difference between the standard deviation with Bessel’s correction and the standard deviation without Bessel’s correction? Which do you think is better and when does it matter?
Bessel’s is better in large populations. Only difference is that Bessel’s does (n-1) to make it more accurate in a certain population size.
Describe the difference between sampling error and sampling bias. Describe the difference between a biased estimator and unbiased estimators.
Sampling bias is the mean sampling error. The sampling error is the difference between the sample mean and the parameter population mean. Unbiased estimator has a sampling error equal to 0.The biased estimator is the difference between the estimator’s expected value and the true vaule of the parameter being estimated.