QQ Plot and Shapiro Wilk Test DisagreeWhat is the meaning of “All models are wrong, but some are useful”Residual plot with a slope of 1What to do when Kolmogorov-Smirnov test is significant for residuals of parametric test but skewness and kurtosis look normal?Heteroskedasticity and residuals normalityLinear regresson lm or stepwise regression here using R?How incorrect is a regression model when assumptions are not met?Residuals analysis: interpretation of a scatter plotNormal Q-Q plot - to check AssumptionHow should I transform the qqplots such that the distribution will be normal?residual plot for positive variableLinear Regression on Boston Housing Price?
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QQ Plot and Shapiro Wilk Test Disagree
What is the meaning of “All models are wrong, but some are useful”Residual plot with a slope of 1What to do when Kolmogorov-Smirnov test is significant for residuals of parametric test but skewness and kurtosis look normal?Heteroskedasticity and residuals normalityLinear regresson lm or stepwise regression here using R?How incorrect is a regression model when assumptions are not met?Residuals analysis: interpretation of a scatter plotNormal Q-Q plot - to check AssumptionHow should I transform the qqplots such that the distribution will be normal?residual plot for positive variableLinear Regression on Boston Housing Price?
$begingroup$
My QQ Plot shows that the data is not normally distributed
qqplot(residual_values, fit = True, line = '45')
pylab.show()
It has a skewness of 0.54
residual_values.skew() # 0.5469389365591185
But the p_value of Shapiro test is greater than 0.05, telling me that it is normally distributed
shapiro(residual_values) # (0.9569438099861145, 0.2261517345905304)
What is the correct inference from this, Is it Normally Distributed or not?
regression machine-learning
asked 5 hours ago
ShinigamiShinigami
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
My QQ Plot shows that the data is not normally distributed
qqplot(residual_values, fit = True, line = '45')
pylab.show()
It has a skewness of 0.54
residual_values.skew() # 0.5469389365591185
But the p_value of Shapiro test is greater than 0.05, telling me that it is normally distributed
shapiro(residual_values) # (0.9569438099861145, 0.2261517345905304)
What is the correct inference from this, Is it Normally Distributed or not?
regression machine-learning
asked 5 hours ago
ShinigamiShinigami
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
The QQ plot looks consistent with being normally distributed. Did you expect every point to fall exactly on the line?
$endgroup$
– The Laconic
3 hours ago
3
$begingroup$
It is approximately normally distributed if you are prepared to discount slight skewness. No procedure ever indicates more.
$endgroup$
– Nick Cox
2 hours ago
$begingroup$
@TheLaconic Sorry I am new to ML.
$endgroup$
– Shinigami
2 hours ago
$begingroup$
@NickCox Thank you Sorry reputation is low to upvote you guys
$endgroup$
– Shinigami
2 hours ago
$begingroup$
It's approximately normal, the skewness in the sample is quite mild; this doesn't automatically mean the population is also skewed (though I expect it is). A high p-value on a test of normality doesn't mean that it is normal, only that you couldn't detect whatever population non-normality there was. (The answer to "is it normally distributed" is "no" - unless you generated it to be normal it won't actually be normal -- but why would it have to be?)
$endgroup$
– Glen_b♦
19 mins ago
add a comment |
$begingroup$
My QQ Plot shows that the data is not normally distributed
qqplot(residual_values, fit = True, line = '45')
pylab.show()
It has a skewness of 0.54
residual_values.skew() # 0.5469389365591185
But the p_value of Shapiro test is greater than 0.05, telling me that it is normally distributed
shapiro(residual_values) # (0.9569438099861145, 0.2261517345905304)
What is the correct inference from this, Is it Normally Distributed or not?
regression machine-learning
asked 5 hours ago
ShinigamiShinigami
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
My QQ Plot shows that the data is not normally distributed
qqplot(residual_values, fit = True, line = '45')
pylab.show()
It has a skewness of 0.54
residual_values.skew() # 0.5469389365591185
But the p_value of Shapiro test is greater than 0.05, telling me that it is normally distributed
shapiro(residual_values) # (0.9569438099861145, 0.2261517345905304)
What is the correct inference from this, Is it Normally Distributed or not?
regression machine-learning
regression machine-learning
asked 5 hours ago
ShinigamiShinigami
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
ShinigamiShinigami
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
ShinigamiShinigami
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
ShinigamiShinigami
164
asked 5 hours ago
ShinigamiShinigami
164
164
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shinigami is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
The QQ plot looks consistent with being normally distributed. Did you expect every point to fall exactly on the line?
$endgroup$
– The Laconic
3 hours ago
3
$begingroup$
It is approximately normally distributed if you are prepared to discount slight skewness. No procedure ever indicates more.
$endgroup$
– Nick Cox
2 hours ago
$begingroup$
@TheLaconic Sorry I am new to ML.
$endgroup$
– Shinigami
2 hours ago
$begingroup$
@NickCox Thank you Sorry reputation is low to upvote you guys
$endgroup$
– Shinigami
2 hours ago
$begingroup$
It's approximately normal, the skewness in the sample is quite mild; this doesn't automatically mean the population is also skewed (though I expect it is). A high p-value on a test of normality doesn't mean that it is normal, only that you couldn't detect whatever population non-normality there was. (The answer to "is it normally distributed" is "no" - unless you generated it to be normal it won't actually be normal -- but why would it have to be?)
$endgroup$
– Glen_b♦
19 mins ago
add a comment |
3
$begingroup$
The QQ plot looks consistent with being normally distributed. Did you expect every point to fall exactly on the line?
$endgroup$
– The Laconic
3 hours ago
3
$begingroup$
It is approximately normally distributed if you are prepared to discount slight skewness. No procedure ever indicates more.
$endgroup$
– Nick Cox
2 hours ago
$begingroup$
@TheLaconic Sorry I am new to ML.
$endgroup$
– Shinigami
2 hours ago
$begingroup$
@NickCox Thank you Sorry reputation is low to upvote you guys
$endgroup$
– Shinigami
2 hours ago
$begingroup$
It's approximately normal, the skewness in the sample is quite mild; this doesn't automatically mean the population is also skewed (though I expect it is). A high p-value on a test of normality doesn't mean that it is normal, only that you couldn't detect whatever population non-normality there was. (The answer to "is it normally distributed" is "no" - unless you generated it to be normal it won't actually be normal -- but why would it have to be?)
$endgroup$
– Glen_b♦
19 mins ago
3
3
$begingroup$
The QQ plot looks consistent with being normally distributed. Did you expect every point to fall exactly on the line?
$endgroup$
– The Laconic
3 hours ago
$begingroup$
The QQ plot looks consistent with being normally distributed. Did you expect every point to fall exactly on the line?
$endgroup$
– The Laconic
3 hours ago
3
3
$begingroup$
It is approximately normally distributed if you are prepared to discount slight skewness. No procedure ever indicates more.
$endgroup$
– Nick Cox
2 hours ago
$begingroup$
It is approximately normally distributed if you are prepared to discount slight skewness. No procedure ever indicates more.
$endgroup$
– Nick Cox
2 hours ago
$begingroup$
@TheLaconic Sorry I am new to ML.
$endgroup$
– Shinigami
2 hours ago
$begingroup$
@TheLaconic Sorry I am new to ML.
$endgroup$
– Shinigami
2 hours ago
$begingroup$
@NickCox Thank you Sorry reputation is low to upvote you guys
$endgroup$
– Shinigami
2 hours ago
$begingroup$
@NickCox Thank you Sorry reputation is low to upvote you guys
$endgroup$
– Shinigami
2 hours ago
$begingroup$
It's approximately normal, the skewness in the sample is quite mild; this doesn't automatically mean the population is also skewed (though I expect it is). A high p-value on a test of normality doesn't mean that it is normal, only that you couldn't detect whatever population non-normality there was. (The answer to "is it normally distributed" is "no" - unless you generated it to be normal it won't actually be normal -- but why would it have to be?)
$endgroup$
– Glen_b♦
19 mins ago
$begingroup$
It's approximately normal, the skewness in the sample is quite mild; this doesn't automatically mean the population is also skewed (though I expect it is). A high p-value on a test of normality doesn't mean that it is normal, only that you couldn't detect whatever population non-normality there was. (The answer to "is it normally distributed" is "no" - unless you generated it to be normal it won't actually be normal -- but why would it have to be?)
$endgroup$
– Glen_b♦
19 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The QQ plot is an informal test of normality that can give you some insight into the nature of deviations from normality; for example, whether the distribution has some skew, or fat tails, or whether there are specific observations that deviate from what you would expect from a normal distribution (outliers). The QQ plot can often convince you that the distribution is definitely not normal, but this isn't such a case. Here, the points fall more or less along the line, which is broadly consistent with normality--intuitively, the sort of variation you would expect to see in a small sample.
The Shapiro test is a formal test of normality. I'm not familiar with the shapiro function's output, so I'm not sure which number, if either, is supposed to be the p-value, but if you say it's largish, then we are led to accept the null hypothesis of normality. And this is consistent with what we see qualitatively in the QQ plot.
answered 1 hour ago
The LaconicThe Laconic
1,0741615
$endgroup$
add a comment |
$begingroup$
The q-q is consistent with (not "proving") approximate normality, more or less.
The Shapiro-Wilk is a formal test of normality and as such, it cannot confirm the null hypothesis of normality. The data may be reasonably consistent with normality yet still be from a different nonnormal underlying distribution. Frequentist hypothesis tests, as a general rule, cannot prove a hypothesis, and failure to reject (p>alpha) does not support the null hypothesis.
@The Laconic gave some decent advice to interpret the q-q plot. However, large p-values do not lead you to accept the null hypothesis (therefore, you don't conclude normality based on this test; the best you can do is say insufficient evidence of nonnormality at the a priori chosen alpha level).
answered 1 hour ago
LSCLSC
1707
$endgroup$
add a comment |
$begingroup$
The shapiro-wilk p-value being >0.05 indicates lack of evidence to against normality. That is consistent with the QQ plot you showed, which is not too far off the line. I don't see what the inconsistency is here. Also, you should give a CI for the skewness coefficient.
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
$endgroup$
add a comment |
$begingroup$
My understanding is that, given power issues with normality tests, they are not highly recommended. As a result I don't use them any more, preferring QQ plots (which are recommended in the literature I have seen).
answered 34 mins ago
user54285user54285
413
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The QQ plot is an informal test of normality that can give you some insight into the nature of deviations from normality; for example, whether the distribution has some skew, or fat tails, or whether there are specific observations that deviate from what you would expect from a normal distribution (outliers). The QQ plot can often convince you that the distribution is definitely not normal, but this isn't such a case. Here, the points fall more or less along the line, which is broadly consistent with normality--intuitively, the sort of variation you would expect to see in a small sample.
The Shapiro test is a formal test of normality. I'm not familiar with the shapiro function's output, so I'm not sure which number, if either, is supposed to be the p-value, but if you say it's largish, then we are led to accept the null hypothesis of normality. And this is consistent with what we see qualitatively in the QQ plot.
answered 1 hour ago
The LaconicThe Laconic
1,0741615
$endgroup$
add a comment |
$begingroup$
The QQ plot is an informal test of normality that can give you some insight into the nature of deviations from normality; for example, whether the distribution has some skew, or fat tails, or whether there are specific observations that deviate from what you would expect from a normal distribution (outliers). The QQ plot can often convince you that the distribution is definitely not normal, but this isn't such a case. Here, the points fall more or less along the line, which is broadly consistent with normality--intuitively, the sort of variation you would expect to see in a small sample.
The Shapiro test is a formal test of normality. I'm not familiar with the shapiro function's output, so I'm not sure which number, if either, is supposed to be the p-value, but if you say it's largish, then we are led to accept the null hypothesis of normality. And this is consistent with what we see qualitatively in the QQ plot.
answered 1 hour ago
The LaconicThe Laconic
1,0741615
$endgroup$
add a comment |
$begingroup$
The QQ plot is an informal test of normality that can give you some insight into the nature of deviations from normality; for example, whether the distribution has some skew, or fat tails, or whether there are specific observations that deviate from what you would expect from a normal distribution (outliers). The QQ plot can often convince you that the distribution is definitely not normal, but this isn't such a case. Here, the points fall more or less along the line, which is broadly consistent with normality--intuitively, the sort of variation you would expect to see in a small sample.
The Shapiro test is a formal test of normality. I'm not familiar with the shapiro function's output, so I'm not sure which number, if either, is supposed to be the p-value, but if you say it's largish, then we are led to accept the null hypothesis of normality. And this is consistent with what we see qualitatively in the QQ plot.
answered 1 hour ago
The LaconicThe Laconic
1,0741615
$endgroup$
The QQ plot is an informal test of normality that can give you some insight into the nature of deviations from normality; for example, whether the distribution has some skew, or fat tails, or whether there are specific observations that deviate from what you would expect from a normal distribution (outliers). The QQ plot can often convince you that the distribution is definitely not normal, but this isn't such a case. Here, the points fall more or less along the line, which is broadly consistent with normality--intuitively, the sort of variation you would expect to see in a small sample.
The Shapiro test is a formal test of normality. I'm not familiar with the shapiro function's output, so I'm not sure which number, if either, is supposed to be the p-value, but if you say it's largish, then we are led to accept the null hypothesis of normality. And this is consistent with what we see qualitatively in the QQ plot.
answered 1 hour ago
The LaconicThe Laconic
1,0741615
answered 1 hour ago
The LaconicThe Laconic
1,0741615
answered 1 hour ago
The LaconicThe Laconic
1,0741615
answered 1 hour ago
The LaconicThe Laconic
1,0741615
1,0741615
add a comment |
add a comment |
$begingroup$
The q-q is consistent with (not "proving") approximate normality, more or less.
The Shapiro-Wilk is a formal test of normality and as such, it cannot confirm the null hypothesis of normality. The data may be reasonably consistent with normality yet still be from a different nonnormal underlying distribution. Frequentist hypothesis tests, as a general rule, cannot prove a hypothesis, and failure to reject (p>alpha) does not support the null hypothesis.
@The Laconic gave some decent advice to interpret the q-q plot. However, large p-values do not lead you to accept the null hypothesis (therefore, you don't conclude normality based on this test; the best you can do is say insufficient evidence of nonnormality at the a priori chosen alpha level).
answered 1 hour ago
LSCLSC
1707
$endgroup$
add a comment |
$begingroup$
The q-q is consistent with (not "proving") approximate normality, more or less.
The Shapiro-Wilk is a formal test of normality and as such, it cannot confirm the null hypothesis of normality. The data may be reasonably consistent with normality yet still be from a different nonnormal underlying distribution. Frequentist hypothesis tests, as a general rule, cannot prove a hypothesis, and failure to reject (p>alpha) does not support the null hypothesis.
@The Laconic gave some decent advice to interpret the q-q plot. However, large p-values do not lead you to accept the null hypothesis (therefore, you don't conclude normality based on this test; the best you can do is say insufficient evidence of nonnormality at the a priori chosen alpha level).
answered 1 hour ago
LSCLSC
1707
$endgroup$
add a comment |
$begingroup$
The q-q is consistent with (not "proving") approximate normality, more or less.
The Shapiro-Wilk is a formal test of normality and as such, it cannot confirm the null hypothesis of normality. The data may be reasonably consistent with normality yet still be from a different nonnormal underlying distribution. Frequentist hypothesis tests, as a general rule, cannot prove a hypothesis, and failure to reject (p>alpha) does not support the null hypothesis.
@The Laconic gave some decent advice to interpret the q-q plot. However, large p-values do not lead you to accept the null hypothesis (therefore, you don't conclude normality based on this test; the best you can do is say insufficient evidence of nonnormality at the a priori chosen alpha level).
answered 1 hour ago
LSCLSC
1707
$endgroup$
The q-q is consistent with (not "proving") approximate normality, more or less.
The Shapiro-Wilk is a formal test of normality and as such, it cannot confirm the null hypothesis of normality. The data may be reasonably consistent with normality yet still be from a different nonnormal underlying distribution. Frequentist hypothesis tests, as a general rule, cannot prove a hypothesis, and failure to reject (p>alpha) does not support the null hypothesis.
@The Laconic gave some decent advice to interpret the q-q plot. However, large p-values do not lead you to accept the null hypothesis (therefore, you don't conclude normality based on this test; the best you can do is say insufficient evidence of nonnormality at the a priori chosen alpha level).
answered 1 hour ago
LSCLSC
1707
answered 1 hour ago
LSCLSC
1707
answered 1 hour ago
LSCLSC
1707
answered 1 hour ago
LSCLSC
1707
1707
add a comment |
add a comment |
$begingroup$
The shapiro-wilk p-value being >0.05 indicates lack of evidence to against normality. That is consistent with the QQ plot you showed, which is not too far off the line. I don't see what the inconsistency is here. Also, you should give a CI for the skewness coefficient.
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
$endgroup$
add a comment |
$begingroup$
The shapiro-wilk p-value being >0.05 indicates lack of evidence to against normality. That is consistent with the QQ plot you showed, which is not too far off the line. I don't see what the inconsistency is here. Also, you should give a CI for the skewness coefficient.
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
$endgroup$
add a comment |
$begingroup$
The shapiro-wilk p-value being >0.05 indicates lack of evidence to against normality. That is consistent with the QQ plot you showed, which is not too far off the line. I don't see what the inconsistency is here. Also, you should give a CI for the skewness coefficient.
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
$endgroup$
The shapiro-wilk p-value being >0.05 indicates lack of evidence to against normality. That is consistent with the QQ plot you showed, which is not too far off the line. I don't see what the inconsistency is here. Also, you should give a CI for the skewness coefficient.
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
answered 1 hour ago
beta1_equals_beta2beta1_equals_beta2
512
512
add a comment |
add a comment |
$begingroup$
My understanding is that, given power issues with normality tests, they are not highly recommended. As a result I don't use them any more, preferring QQ plots (which are recommended in the literature I have seen).
answered 34 mins ago
user54285user54285
413
$endgroup$
add a comment |
$begingroup$
My understanding is that, given power issues with normality tests, they are not highly recommended. As a result I don't use them any more, preferring QQ plots (which are recommended in the literature I have seen).
answered 34 mins ago
user54285user54285
413
$endgroup$
add a comment |
$begingroup$
My understanding is that, given power issues with normality tests, they are not highly recommended. As a result I don't use them any more, preferring QQ plots (which are recommended in the literature I have seen).
answered 34 mins ago
user54285user54285
413
$endgroup$
My understanding is that, given power issues with normality tests, they are not highly recommended. As a result I don't use them any more, preferring QQ plots (which are recommended in the literature I have seen).
answered 34 mins ago
user54285user54285
413
answered 34 mins ago
user54285user54285
413
answered 34 mins ago
user54285user54285
413
answered 34 mins ago
user54285user54285
413
413
add a comment |
add a comment |
Shinigami is a new contributor. Be nice, and check out our Code of Conduct.
Shinigami is a new contributor. Be nice, and check out our Code of Conduct.
Shinigami is a new contributor. Be nice, and check out our Code of Conduct.
Shinigami is a new contributor. Be nice, and check out our Code of Conduct.
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3
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The QQ plot looks consistent with being normally distributed. Did you expect every point to fall exactly on the line?
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– The Laconic
3 hours ago
3
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It is approximately normally distributed if you are prepared to discount slight skewness. No procedure ever indicates more.
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– Nick Cox
2 hours ago
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@TheLaconic Sorry I am new to ML.
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– Shinigami
2 hours ago
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@NickCox Thank you Sorry reputation is low to upvote you guys
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– Shinigami
2 hours ago
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It's approximately normal, the skewness in the sample is quite mild; this doesn't automatically mean the population is also skewed (though I expect it is). A high p-value on a test of normality doesn't mean that it is normal, only that you couldn't detect whatever population non-normality there was. (The answer to "is it normally distributed" is "no" - unless you generated it to be normal it won't actually be normal -- but why would it have to be?)
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– Glen_b♦
19 mins ago