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Rom4ik [11]
1 year ago
11

the correlation between two variables related to the same sample is a scaled quantitative measure. what is used to scale this me

asurement?
Mathematics
1 answer:
Doss [256]1 year ago
6 0

The correct option is the standard deviations of the variables.

What is correlation?

Any statistical relationship between two random variables or bivariate data, whether causal or not, is referred to in statistics as correlation or dependence.

Correlation is a term used to describe the linear relationship between two entities. When using a correlation tool, the variables must be present in quantities of two or more to test their relationship (degree of association).

In order to scale the measurement between two variables connected to a similar sample, the standard deviation measures dispersion.

The variables' standard deviations are the best choice as a result.

Option involving the standard errors of the variables is incorrect because standard error, which stands for standard deviation, determines the relationship between dependent and independent variables.

Option involving the coefficients of variation of two variables is incorrect because the coefficient expressing standard deviation and mean ratio determines how much variation there is.

Option The means of the variables are incorrect because they are not a scale for determining the correlation between two variables.

Hence, the correct option is the standard deviations of the variables.

To know more about correlation, click on the link

brainly.com/question/2088651

#SPJ4

Complete question:

The correlation between two variables related to the same sample is a scaled quantitative measure. What is used to scale this measurement?

- the standard deviations of the variables

- the standard errors of the variables

- the coefficients of variation of the two variables

- the means of the variables

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Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 3

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We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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