Is it correct to say that a z-value follows a standard normal distribution?
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As the title reveals; is it correct to say that a z-value follows a standard normal distribution? The sentence that I use it in, is as follows:
Moreover, all p-values are reported with the corresponding test statistic: the z-value that follows a standard normal distribution.
hypothesis-testing statistical-significance normal-distribution
asked Aug 23 at 17:32
Amonet
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up vote
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As the title reveals; is it correct to say that a z-value follows a standard normal distribution? The sentence that I use it in, is as follows:
Moreover, all p-values are reported with the corresponding test statistic: the z-value that follows a standard normal distribution.
hypothesis-testing statistical-significance normal-distribution
asked Aug 23 at 17:32
Amonet
15811
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
As the title reveals; is it correct to say that a z-value follows a standard normal distribution? The sentence that I use it in, is as follows:
Moreover, all p-values are reported with the corresponding test statistic: the z-value that follows a standard normal distribution.
hypothesis-testing statistical-significance normal-distribution
asked Aug 23 at 17:32
Amonet
15811
As the title reveals; is it correct to say that a z-value follows a standard normal distribution? The sentence that I use it in, is as follows:
Moreover, all p-values are reported with the corresponding test statistic: the z-value that follows a standard normal distribution.
hypothesis-testing statistical-significance normal-distribution
hypothesis-testing statistical-significance normal-distribution
asked Aug 23 at 17:32
Amonet
15811
asked Aug 23 at 17:32
Amonet
15811
asked Aug 23 at 17:32
Amonet
15811
asked Aug 23 at 17:32
Amonet
15811
asked Aug 23 at 17:32
Amonet
15811
15811
add a comment |Â
add a comment |Â
1 Answer
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Not necessarily. Sometimes the term "z value" is used to describe a standardized variable $Z$ (or a single value $z$), which is a set of observed values (realizations of some distribution) with a population mean of exactly 0, and a population standard deviation of exactly 1 derived from another variable $X$ through the transformation:
$$Z = fracX-mu_Xsigma_X$$
If the distribution of $X$ is normal, then $Z$ will have a standard normal distribution. However, if $X$ is not normal, then $Z$ will not be either, even though it still has a population mean of exactly 0, and population standard deviation of exactly 1 (i.e. $mu_Z = 0, sigma_Z=1$).
answered Aug 23 at 17:51
Alexis
15.4k34494
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Not necessarily. Sometimes the term "z value" is used to describe a standardized variable $Z$ (or a single value $z$), which is a set of observed values (realizations of some distribution) with a population mean of exactly 0, and a population standard deviation of exactly 1 derived from another variable $X$ through the transformation:
$$Z = fracX-mu_Xsigma_X$$
If the distribution of $X$ is normal, then $Z$ will have a standard normal distribution. However, if $X$ is not normal, then $Z$ will not be either, even though it still has a population mean of exactly 0, and population standard deviation of exactly 1 (i.e. $mu_Z = 0, sigma_Z=1$).
answered Aug 23 at 17:51
Alexis
15.4k34494
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
add a comment |Â
up vote
8
down vote
accepted
Not necessarily. Sometimes the term "z value" is used to describe a standardized variable $Z$ (or a single value $z$), which is a set of observed values (realizations of some distribution) with a population mean of exactly 0, and a population standard deviation of exactly 1 derived from another variable $X$ through the transformation:
$$Z = fracX-mu_Xsigma_X$$
If the distribution of $X$ is normal, then $Z$ will have a standard normal distribution. However, if $X$ is not normal, then $Z$ will not be either, even though it still has a population mean of exactly 0, and population standard deviation of exactly 1 (i.e. $mu_Z = 0, sigma_Z=1$).
answered Aug 23 at 17:51
Alexis
15.4k34494
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Not necessarily. Sometimes the term "z value" is used to describe a standardized variable $Z$ (or a single value $z$), which is a set of observed values (realizations of some distribution) with a population mean of exactly 0, and a population standard deviation of exactly 1 derived from another variable $X$ through the transformation:
$$Z = fracX-mu_Xsigma_X$$
If the distribution of $X$ is normal, then $Z$ will have a standard normal distribution. However, if $X$ is not normal, then $Z$ will not be either, even though it still has a population mean of exactly 0, and population standard deviation of exactly 1 (i.e. $mu_Z = 0, sigma_Z=1$).
answered Aug 23 at 17:51
Alexis
15.4k34494
Not necessarily. Sometimes the term "z value" is used to describe a standardized variable $Z$ (or a single value $z$), which is a set of observed values (realizations of some distribution) with a population mean of exactly 0, and a population standard deviation of exactly 1 derived from another variable $X$ through the transformation:
$$Z = fracX-mu_Xsigma_X$$
If the distribution of $X$ is normal, then $Z$ will have a standard normal distribution. However, if $X$ is not normal, then $Z$ will not be either, even though it still has a population mean of exactly 0, and population standard deviation of exactly 1 (i.e. $mu_Z = 0, sigma_Z=1$).
answered Aug 23 at 17:51
Alexis
15.4k34494
answered Aug 23 at 17:51
Alexis
15.4k34494
answered Aug 23 at 17:51
Alexis
15.4k34494
answered Aug 23 at 17:51
Alexis
15.4k34494
15.4k34494
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
add a comment |Â
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
Thanks for the elaborate explanation. How would you advice to report it instead? My study's context is confirmatory factor analysis (within structural equation modelling), where the assumption is being made that each observed / measured variable is normally distributed. This z-value is reported in the output with every significance test - with the null being that the estimate is zero (I think this is similar to OLS). Btw: I can update my question with this part, if that is preferred.
– Amonet
Aug 23 at 18:00
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
You are welcome! (Also: Welcome to CV!) Ask a new question! :) Although, given "the assumption is being made that each observed / measured variable is normally distributed," do you actually have $mu_X$ and $sigma_X$ (population statistics)? Or do you just have $barX$ and $s_X$ (sample statistics)? If the latter you do not have a $Z$ variable, but a $T$ variable: $T=fracX-barXs_X$.
– Alexis
Aug 23 at 18:07
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
I looked into the documentation of the R package (Lavaan), and it says the z-value is the test-statistic of the Wald test. The z-value is simply the parameter estimate divided by the standard error. While I understand what you're saying, I believe the R package is simply giving me the Z variable. Otherwise I think it would have reported P(>|t|) for example instead of P(>|z|) and it would have given me t-values rather than z-values. So, I am sure you're correct that it should be a T variable, but it's giving me a Z variable. Do you agree?
– Amonet
Aug 23 at 18:22
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
"The z-value is simply the parameter estimate divided by the standard error" says nothing about whether that is a sample or population standard error. I am unfamiliar with the package. The authors may have made a mistake, or may be assuming things that make their statement true given the assumptions. Under the null hypothesis, the population mean and variance of Bernouli data are known, so $Z$ can be calculated. You should post another question, rather than trying to get new questions answered in the comments. :)
– Alexis
Aug 23 at 18:25
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
Sorry, you are right. Thanks again for helping me out so well!
– Amonet
Aug 23 at 18:31
add a comment |Â
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