Recent questions in Inferential Statistics

Inferential StatisticsAnswered question

Ayanna Jarvis 2022-10-12

Find the upper estimate for the total amount of water that leaked out by using five rectangles? Give your answer with one decimal place.

Inferential StatisticsAnswered question

Janessa Benson 2022-10-07

Probability of infection by staphylococcus aureus

Please forgive my innumeracy, but I have a question with which I am hoping someone might be able to help me.

Suppose the following be true. The chance of a prosthetic hip joint becoming infected by staphylococcus aureus is one per cent. The chance of a natural hip joint becoming infected by staphylococcus aureus is 0.1%. In other words (in case I am misusing the word 'chance'), one in one hundred people with prosthetic hip joints will become infected by staphylococcus aureus, whereas only one in one thousand people with natural hip joints will become so infected.

Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)? In other words (to make it clear what I mean by 'probably'), is it correct to say that there is a greater than 50 per cent chance that X's hip joint would not have become infected but for the fact that X has a prosthetic hip joint? Why or why not?

Please forgive my innumeracy, but I have a question with which I am hoping someone might be able to help me.

Suppose the following be true. The chance of a prosthetic hip joint becoming infected by staphylococcus aureus is one per cent. The chance of a natural hip joint becoming infected by staphylococcus aureus is 0.1%. In other words (in case I am misusing the word 'chance'), one in one hundred people with prosthetic hip joints will become infected by staphylococcus aureus, whereas only one in one thousand people with natural hip joints will become so infected.

Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)? In other words (to make it clear what I mean by 'probably'), is it correct to say that there is a greater than 50 per cent chance that X's hip joint would not have become infected but for the fact that X has a prosthetic hip joint? Why or why not?

Inferential StatisticsAnswered question

Cindy Noble 2022-10-05

How to write an equation where both independent variables and dependent variables are log transformed in a multiple regression?

How to write the multiple regression model when both the dependent variable and independent variables are log-transformed?

I know that without any log transformation the linear regression model would be written as enter image description here

$y={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

But now I have transformed both my dependent variables and independent variable with log. So is correct to write as enter image description here $\mathrm{log}(y)={\beta}_{0}+{\beta}_{1}\cdot \mathrm{log}({x}_{1})+{\beta}_{2}\cdot \mathrm{log}({x}_{2})+\dots $

Or since I am transforming both sides of question so can I write it as enter image description here

$\mathrm{ln}(y)={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

How to write the multiple regression model when both the dependent variable and independent variables are log-transformed?

I know that without any log transformation the linear regression model would be written as enter image description here

$y={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

But now I have transformed both my dependent variables and independent variable with log. So is correct to write as enter image description here $\mathrm{log}(y)={\beta}_{0}+{\beta}_{1}\cdot \mathrm{log}({x}_{1})+{\beta}_{2}\cdot \mathrm{log}({x}_{2})+\dots $

Or since I am transforming both sides of question so can I write it as enter image description here

$\mathrm{ln}(y)={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

Inferential StatisticsAnswered question

Robaffonadorkdh 2022-10-02

Based on the estimates $\mathrm{log}(2)=.03$ and $\mathrm{log}(5)=.7$, how do you use properties of logarithms to find approximate values for ${\mathrm{log}}_{5}(2)$?

Inferential StatisticsAnswered question

Leonel Schwartz 2022-10-02

Please, give examples of two variables that have a perfect positive linear correlation and two variables that have a perfect negative linear correlation.

Inferential StatisticsAnswered question

garnirativ8 2022-09-30

Let us define the correlation coefficient as $\rho (X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$ .

Are the following statements true or false?

If $\rho (X,Y)=\rho (Y,Z)=0$ then $\rho (X,Z)=0$

If $\rho (X,Y)>\rho (Y,Z)>0$ then $\rho (X,Z)>0$

If $\rho (X,Y)<\rho (Y,Z)<0$ then $\rho (X,Z)<0$

I think they are false, but I can't find counterexamples. Could you help me?

Are the following statements true or false?

If $\rho (X,Y)=\rho (Y,Z)=0$ then $\rho (X,Z)=0$

If $\rho (X,Y)>\rho (Y,Z)>0$ then $\rho (X,Z)>0$

If $\rho (X,Y)<\rho (Y,Z)<0$ then $\rho (X,Z)<0$

I think they are false, but I can't find counterexamples. Could you help me?

Inferential StatisticsAnswered question

ecoanuncios7x 2022-09-30

Linear Regression:

$$Y=a+bX+\u03f5$$

For $$R$$ squared in linear regression, in the form of ratio between $({y}_{i}-{y}^{bar})$, or in terms of

$$({S}_{xy}{)}^{2}/({S}_{xx}{S}_{yy})$$

Not sure if you guys come across this form:

$${R}^{2}=\frac{Var(bX)}{V(bX)+V(\u03f5)}$$?

$$Y=a+bX+\u03f5$$

For $$R$$ squared in linear regression, in the form of ratio between $({y}_{i}-{y}^{bar})$, or in terms of

$$({S}_{xy}{)}^{2}/({S}_{xx}{S}_{yy})$$

Not sure if you guys come across this form:

$${R}^{2}=\frac{Var(bX)}{V(bX)+V(\u03f5)}$$?

Inferential StatisticsAnswered question

Aidyn Crosby 2022-09-29

A and B have negative correlation, so -A and -B have positive?

Inferential StatisticsAnswered question

Aidyn Crosby 2022-09-29

What is the main difference between correlation and causation? (Answer the question in a short paragraph (roughly 3-5 sentences). If necessary, explain the concept and/or give examples)

Inferential StatisticsAnswered question

elisegayezm 2022-09-29

Based on the estimates $\mathrm{log}(2)=.03$ and $\mathrm{log}(5)=.7$, how do you use properties of logarithms to find approximate values for $\mathrm{log}(0.25)$?

Inferential StatisticsAnswered question

eukrasicx 2022-09-29

Proof that given 2 variables X and Y with correlation ${\rho}_{X,Y}$ and given $U=a+bX$ and $V=c+dY$ then ${\rho}_{X,Y}={\rho}_{U,V}$ if $bd>0$

Inferential StatisticsAnswered question

garnirativ8 2022-09-28

How do you determine the value of an x-ray machine after 5 year if it cost $216 thousand and Margaret Madison, DDS, estimates that her dental equipment loses one sixth of its value each year?

Inferential StatisticsAnswered question

Mangle Woods2022-09-27

what ia the probability of obtaining ten heads in a row when flipping a coin? Interpret this probability.

the probability of obtaining ten heads in a row when flipping a coin it?

Inferential StatisticsAnswered question

Parker Pitts 2022-09-27

Multiple Regression Forecast

"Part C: asks what salary would you forecast for a man with 12 years of education, 10 months of experience, and 15 months with the company."

This is straight forward enough just reading off the coefficients table. $y=3526.4+(722.5)(1)+(90.02)(12)+(1.269)(10)+(23.406)(15)=5692.92$

but

"Part D: asks what salary would you forecast for men with 12 years of education, 10 months of experience, and 15 months with the company."

I know that the answer to this must be different from C, but I have no idea why, I would of just done exactly the same as in part C,

What is wrong with my train of thought or intuition and how might I go about calculating the salary for men, rather than a man?

"Part C: asks what salary would you forecast for a man with 12 years of education, 10 months of experience, and 15 months with the company."

This is straight forward enough just reading off the coefficients table. $y=3526.4+(722.5)(1)+(90.02)(12)+(1.269)(10)+(23.406)(15)=5692.92$

but

"Part D: asks what salary would you forecast for men with 12 years of education, 10 months of experience, and 15 months with the company."

I know that the answer to this must be different from C, but I have no idea why, I would of just done exactly the same as in part C,

What is wrong with my train of thought or intuition and how might I go about calculating the salary for men, rather than a man?

Inferential StatisticsAnswered question

trkalo84 2022-09-27

What is $Var[b]$ in multiple regression?

Assume a linear regression model $y=X\beta +\u03f5$ with $\u03f5\sim N(0,{\sigma}^{2}I)$ and $\hat{y}=Xb$ where $b=({X}^{\prime}X{)}^{-1}{X}^{\prime}y$. Besides $H=X({X}^{\prime}X{)}^{-1}{X}^{\prime}$ is the linear projection from the response space to the span of $X$, i.e., $\hat{y}=Hy$

Now I want to calculate $Var[b]$ but what I get is an $k\times k$ matrix, not an $n\times n$ one. Here's my calculation:

$\begin{array}{rl}Var[b]=& \phantom{\rule{thickmathspace}{0ex}}Var[({X}^{\prime}X{)}^{-1}{X}^{\prime}y]\\ =& \phantom{\rule{thickmathspace}{0ex}}({X}^{\prime}X{)}^{-1}{X}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\underset{={\sigma}^{2}I}{\underset{\u23df}{Var[y]}}X({X}^{\prime}X{)}^{-1}\\ \\ \text{Here you can}& \text{see already this thing will be k}\times \text{k}\\ \\ =& \phantom{\rule{thickmathspace}{0ex}}{\sigma}^{2}\underset{I}{\underset{\u23df}{({X}^{\prime}X{)}^{-1}{X}^{\prime}X}}({X}^{\prime}X{)}^{-1}\\ =& {\sigma}^{2}({X}^{\prime}X{)}^{-1}\phantom{\rule{thinmathspace}{0ex}}\in {R}^{k\times k}\end{array}$

What am I doing wrong?

Besides, are $E[b]=\beta $, $E[\hat{y}]=HX\beta $, $Var[\hat{y}]={\sigma}^{2}H$, $E[y-\hat{y}]=(I-H)X\beta $, $Var[y-\hat{y}]=(I-H){\sigma}^{2}$ correct (this is just on a side note, my main question is the one above)?

Assume a linear regression model $y=X\beta +\u03f5$ with $\u03f5\sim N(0,{\sigma}^{2}I)$ and $\hat{y}=Xb$ where $b=({X}^{\prime}X{)}^{-1}{X}^{\prime}y$. Besides $H=X({X}^{\prime}X{)}^{-1}{X}^{\prime}$ is the linear projection from the response space to the span of $X$, i.e., $\hat{y}=Hy$

Now I want to calculate $Var[b]$ but what I get is an $k\times k$ matrix, not an $n\times n$ one. Here's my calculation:

$\begin{array}{rl}Var[b]=& \phantom{\rule{thickmathspace}{0ex}}Var[({X}^{\prime}X{)}^{-1}{X}^{\prime}y]\\ =& \phantom{\rule{thickmathspace}{0ex}}({X}^{\prime}X{)}^{-1}{X}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\underset{={\sigma}^{2}I}{\underset{\u23df}{Var[y]}}X({X}^{\prime}X{)}^{-1}\\ \\ \text{Here you can}& \text{see already this thing will be k}\times \text{k}\\ \\ =& \phantom{\rule{thickmathspace}{0ex}}{\sigma}^{2}\underset{I}{\underset{\u23df}{({X}^{\prime}X{)}^{-1}{X}^{\prime}X}}({X}^{\prime}X{)}^{-1}\\ =& {\sigma}^{2}({X}^{\prime}X{)}^{-1}\phantom{\rule{thinmathspace}{0ex}}\in {R}^{k\times k}\end{array}$

What am I doing wrong?

Besides, are $E[b]=\beta $, $E[\hat{y}]=HX\beta $, $Var[\hat{y}]={\sigma}^{2}H$, $E[y-\hat{y}]=(I-H)X\beta $, $Var[y-\hat{y}]=(I-H){\sigma}^{2}$ correct (this is just on a side note, my main question is the one above)?

Inferential StatisticsAnswered question

Melina Barber 2022-09-26

Pearson Correlation Coefficient Interpretation

Let X=(1,2,3,...,20). Suppose that $Y=({y}_{1},{y}_{2},...,{y}_{20})$ with ${y}_{i}={x}_{i}^{2}$ and $Z=({z}_{1},{z}_{2},...,{z}_{20})$ with ${z}_{i}={e}^{{x}_{i}}$. Pearson correlation coefficient is defined by formula

$\rho (X,Y)=\frac{\sum _{i=1}^{20}({x}_{i}-\overline{x})({y}_{i}-\overline{y})}{\sqrt{(\sum _{i=1}^{20}({x}_{i}-\overline{x}{)}^{2})(\sum _{i=1}^{20}({y}_{i}-\overline{y}{)}^{2})}}$

If $\rho (X,Y)=1$, we can say that X and Y have a linear correlation. If $0.7\le \rho (X,Y)<1$ then X and Y has a strong linear correlation, if $0.5\le \rho (X,Y)<0.7$ then X and Y has a modest linear correlation, and if $0\le \rho (X,Y)<0.5$ then X and Y has a weak linear correlation. Using this formula, we get $\rho $(X,Y)=0.9 and $\rho $(X,Z)=0.5. However, the relationship between X and Y is actually quadratic but they have the high correlation coefficient that indicate linear correlation.

So, my question is what is "linear correlation" actually between X and Y ? Since $\rho $(X,Z)=0.5 indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\rho $(X,Y) and $\rho $(X,Z), noting that Y and Z is not a linear function of X.

Let X=(1,2,3,...,20). Suppose that $Y=({y}_{1},{y}_{2},...,{y}_{20})$ with ${y}_{i}={x}_{i}^{2}$ and $Z=({z}_{1},{z}_{2},...,{z}_{20})$ with ${z}_{i}={e}^{{x}_{i}}$. Pearson correlation coefficient is defined by formula

$\rho (X,Y)=\frac{\sum _{i=1}^{20}({x}_{i}-\overline{x})({y}_{i}-\overline{y})}{\sqrt{(\sum _{i=1}^{20}({x}_{i}-\overline{x}{)}^{2})(\sum _{i=1}^{20}({y}_{i}-\overline{y}{)}^{2})}}$

If $\rho (X,Y)=1$, we can say that X and Y have a linear correlation. If $0.7\le \rho (X,Y)<1$ then X and Y has a strong linear correlation, if $0.5\le \rho (X,Y)<0.7$ then X and Y has a modest linear correlation, and if $0\le \rho (X,Y)<0.5$ then X and Y has a weak linear correlation. Using this formula, we get $\rho $(X,Y)=0.9 and $\rho $(X,Z)=0.5. However, the relationship between X and Y is actually quadratic but they have the high correlation coefficient that indicate linear correlation.

So, my question is what is "linear correlation" actually between X and Y ? Since $\rho $(X,Z)=0.5 indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\rho $(X,Y) and $\rho $(X,Z), noting that Y and Z is not a linear function of X.

Inferential StatisticsAnswered question

malaana5k 2022-09-26

Meaning of 'obligatory disclaimer'

Can you explain to me what 'obligatory disclaimer' means in probability. Is it like a lack of information or what?

The context is the following:

Second, we found that the marginal distribution of Y is Bern(0.08), whereas the conditional distribution of Y given X = 1 is Bern(0.2) and the conditional distribution of Y given X = 0 is Bern(0.04). Since conditioning on the value of X alters the distribution of Y , X and Y are not independent: learning whether or not the sampled individual is a current smoker gives us information about the probability that he will develop lung cancer. This example comes with an obligatory disclaimer. Although we have found that X and Y are dependent, we cannot make conclusions about whether smoking causes lung cancer based on this association alone. (Joseph K. Blitzstein, Jessica Hwang--Introduction to Probability)

Can you explain to me what 'obligatory disclaimer' means in probability. Is it like a lack of information or what?

The context is the following:

Second, we found that the marginal distribution of Y is Bern(0.08), whereas the conditional distribution of Y given X = 1 is Bern(0.2) and the conditional distribution of Y given X = 0 is Bern(0.04). Since conditioning on the value of X alters the distribution of Y , X and Y are not independent: learning whether or not the sampled individual is a current smoker gives us information about the probability that he will develop lung cancer. This example comes with an obligatory disclaimer. Although we have found that X and Y are dependent, we cannot make conclusions about whether smoking causes lung cancer based on this association alone. (Joseph K. Blitzstein, Jessica Hwang--Introduction to Probability)

Inferential StatisticsAnswered question

Jazmyn Pugh 2022-09-25

Unable to understand correlation coefficient / auto correlation

Suppose I have a vector say:

[5 5 5 5 4 5]

then common sense says that there is a very high auto-correlation for the vector because it is more or less the same values. But when I try to calculate the auto-correlation coefficient, I'm getting a very low value(<0.3) for all lags. What does this mean? shouldn't it be higher because the series is very similar?Am I missing something?does correlatiom mean not similarity but similarity in rate(Rate of change)?

Suppose I have a vector say:

[5 5 5 5 4 5]

then common sense says that there is a very high auto-correlation for the vector because it is more or less the same values. But when I try to calculate the auto-correlation coefficient, I'm getting a very low value(<0.3) for all lags. What does this mean? shouldn't it be higher because the series is very similar?Am I missing something?does correlatiom mean not similarity but similarity in rate(Rate of change)?

Inferential StatisticsAnswered question

deiluefniwf 2022-09-24

The formula that I know for correlation coefficient

$\frac{\sum ({x}_{i}-\overline{x})({Y}_{i}-\overline{Y})}{\sqrt{\sum ({x}_{i}-\overline{x}{)}^{2}\sum ({Y}_{i}-\overline{Y}{)}^{2}}}$

If the only given values I have are $\sum {x}_{i},\sum {x}_{i}^{2},\sum {y}_{i},\sum {x}_{i}{y}_{i}$ is it even possible to compute the correlation coefficient?

$\frac{\sum ({x}_{i}-\overline{x})({Y}_{i}-\overline{Y})}{\sqrt{\sum ({x}_{i}-\overline{x}{)}^{2}\sum ({Y}_{i}-\overline{Y}{)}^{2}}}$

If the only given values I have are $\sum {x}_{i},\sum {x}_{i}^{2},\sum {y}_{i},\sum {x}_{i}{y}_{i}$ is it even possible to compute the correlation coefficient?

In simple terms, inferential statistics is an approach where you use measurements from the sample of specific subjects as you conduct an experiment. The purpose is to make an outcome based on generalization regarding the greater population of subjects. You may use equations if there are questions that are related to a particular approach. You can get inferential statistics help as we provide a list of answers with good samples to start with. There are related topics like correlation problems that will help you with financial statistics and the coordination of variables