Statistics Test - 55

Thestandard error (SE)of a statistic is theapproximate standard deviation of a statistical sample population.

The standard error is a statistical term that measures the accuracy with which a sample distribution represents a population by using standard deviation.

In statistics,a sample mean deviates from the actual mean of a population; this deviation is the standard error of the mean.

Standard deviation (SD) and the estimated standard error of the mean (SEM) are used to present the characteristics of sample data and to explain statistical analysis results.

For a sample size 'n' and\(\bar{x}\),

The standard deviation (SD) is given by:

\(\sigma=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n-1}}\)

Variance \(=\sigma^{2}\)

Now the standard error is given by:

\(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

From the above formula, it is clear that the Standard error of the sample mean is the ratio of standard deviation and the square root of the number of observations.

Given:

Pearsaons measure of skewness \(=0.42\)

Mean \(=\bar{x}=86\)

Median \(=M_{d}=80\)

We know that:

Skewness \((Sk_{p})=\) \(\frac{3(\text{Mean}-\text{median})}{\text{standard deviation}}\)

\(0.42=\frac{3\left(\overline{ x }- M _{ d }\right)}{ \sigma}\)

\(\Rightarrow 0.42=\frac{3(86-80)}{\sigma}\)

\(\Rightarrow \sigma=\frac{18}{0.42}\)

\(\Rightarrow \sigma=\frac{300}{7}\)

Coefficient of variation \(=\) standard

CV \(=\left(\frac{\frac{300}{7}}{86}\right) \times 100\)

\(\Rightarrow \left(\frac{300}{7} \times 86\right) \times 100\)

\(\therefore\) Coefficient of variation is 50.

Median of 'n' observations is the value at the central position when the data is arranged in ascending or descending order.

For, odd \(n\), median is the value at the \(\left(\frac{n+1}{2}\right)\) position.

For even \(n\), median is the mean of values at \(\left(\frac{n}{2}\right)\) and \(\left(\frac{n}{2}+1\right)\) position.

For the given set, number of observations, \(n=6\) (even).

\(\therefore\) Median will be the mean of the values at \(\left(\frac{6}{2}\right)=\) 3rd and \(\left(\frac{6}{2}+1\right)=3+1=4\)th position.

\(\Rightarrow \) Median \(=\frac{(x+3)+(x+4)}{2}=x+3.5\)

Given:

Pearsaons measure of skewness \(=0.33\)

Mean \((\bar{x})=116\)

Median \(\left(M_{d}\right)=105\)

We know that,

Skewness \(\left( Sk _{ p }\right)=\frac{3(\text {Mean}-\text {Median})}{\text { Standard deviation }}\)

\(0.33=\frac{3\left(\overline{ x }- M _{ d }\right)}{\sigma}\)

\(\Rightarrow 0.33=\frac{3(116-105)}{\sigma}\)

\(\Rightarrow \sigma=\frac{33}{0.33}\)

\(\Rightarrow \sigma=100\)

Then, the standard deviation is 100.

Coefficient of variation \(( CV )=\frac{\text { Standard deviation }}{\text { Mean }}\)

\(\left(\frac{\sigma}{\bar{x}}\right) \times 100\)

\(\Rightarrow CV =\left(\frac{100}{116}\right) \times 100\)

\(\Rightarrow(0.8621) \times 100\)

\(\therefore\) Coefficient of variation is \(86.21\).

Given: Ina frequency distribution, the mean and median are 10 and 9 respectively

Relationship between Mean, Median and mode:

As we know that,Mode = 3 Median – 2 Mean

Mode = (3× 9) - (2× 10) = 7

We know that:

Mean \(=\sum \frac{x_{i}}{N}\)

\(G M=N \sqrt{x}_{1} \times x_{2} \times \ldots \ldots x_{N}\)

\(H M=\frac{N}{\Sigma}(\frac{1}{x_{i}})\)

Where, N is the number of observations

Let, us take an observation \(4,6,2\)

Arithmetic mean \(=\frac{(4+6+2)}{3}\)

\(\therefore\) Mean \(=\frac{12}{3}=4\)

Harmonic mean \(=\) It is reciprocal of AM

\(\therefore HM =\frac{1}{4}\)

\(GM =\sqrt{ (AM \times HM )}\)

\({ GM =\sqrt{\left(4 \times \frac{1}{4}\right)}}\)

\({\therefore GM =1}\)

From this result we can say that \(AM > GM > HM\)

When we take some other observation which follows \(AM = GM = HM\)

\(\therefore\) The relation between \(AM , GM\), and \(HM\) is \(AM \geq GM \geq HM\)

Given:

Arithmetic mean \(=(\mu)=16\)

variance \(=\sigma^{2}=4\)

We know that:

Standard deviation \(=\sigma=\sqrt{\text { variance }}\)

\(C V=\) Coefficient of variation \(=\frac{\sigma}{\mu}\)

\(\sigma=\) Standard deviation

\(\mu=\) Mean

Standard deviation \(=\sigma=\sqrt{4}=2\)

\(CV =\left(\frac{\sigma}{\mu}\right) \times 100\)

\(\Rightarrow C V=\left(\frac{2}{16}\right) \times 100\)

\(\therefore\) Coefficient of variation is \(12.5\).

As per the given data,

Mean (\(\bar{x}\)) of \((18,23,14,3,17)= \frac{(18+23+14+3+17)}{5}=15\)

Variance

\(\sigma^{2}=\frac{1}{n} \times \Sigma\left(x_{i}-\bar{x}\right)^{2}\)

\(=\frac{\left[(18-15)^{2}+(23-15)^{2}+(14-15)^{2}+(3-15)^{2}+(17-15)^{2}\right]}{5}\)

\(=\frac{(9+64+1+144+4)}{5}\)

\(=\frac{222}{5}\)

\(=44.4\)

Standard Deviation: If \(\sigma^{2}\) is the variance, then \(\sigma\), is called the standard deviation, is given by

\(\sigma=\sqrt{\frac{1}{n} \times \Sigma\left(x_{i}-\bar{x}\right)^{2}}\)

Standard deviation \(=\sqrt{\text { variance }}=\sqrt{44.4}=6.66\)

Given: The standard deviation and the mean of 16 values are 15.6 and 20.5 respectively

Here, we have to find the coefficient of variation for the same observations.

As we know, Coefficient of variation \(=\frac{\text { standard deviaiton }}{\text { Mean }} \times 100\)

Coefficient of variation \(=\frac{15.6}{20.5} \times 100=76 \%\) (approximately)