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P-value of a variance?
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I know that the p-value is the probability of getting a value more extreme than the one observed given that the null hypothesis is true.
I know that so far we use the z-test and t-test to approximate the mean or calculate the p-value of a mean and the z-test and binomial to approximate or calculate the p-value of a proportion.But how can we calculate the p-value if the parameter is a variance?
probability
asked Jul 28 at 19:54
Roy Rizk
887
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up vote
0
down vote
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I know that the p-value is the probability of getting a value more extreme than the one observed given that the null hypothesis is true.
I know that so far we use the z-test and t-test to approximate the mean or calculate the p-value of a mean and the z-test and binomial to approximate or calculate the p-value of a proportion.But how can we calculate the p-value if the parameter is a variance?
probability
asked Jul 28 at 19:54
Roy Rizk
887
There is an important distinction between (a) normal tables, which have probabilities in the body of the table and cut-off values in the margins and (b) tables of t, chi-squared, and F distributions, which have cutoff values in the body of the table and probabilities in the margins.
– BruceET
Jul 29 at 2:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that the p-value is the probability of getting a value more extreme than the one observed given that the null hypothesis is true.
I know that so far we use the z-test and t-test to approximate the mean or calculate the p-value of a mean and the z-test and binomial to approximate or calculate the p-value of a proportion.But how can we calculate the p-value if the parameter is a variance?
probability
asked Jul 28 at 19:54
Roy Rizk
887
I know that the p-value is the probability of getting a value more extreme than the one observed given that the null hypothesis is true.
I know that so far we use the z-test and t-test to approximate the mean or calculate the p-value of a mean and the z-test and binomial to approximate or calculate the p-value of a proportion.But how can we calculate the p-value if the parameter is a variance?
probability
asked Jul 28 at 19:54
Roy Rizk
887
asked Jul 28 at 19:54
Roy Rizk
887
asked Jul 28 at 19:54
Roy Rizk
887
asked Jul 28 at 19:54
Roy Rizk
887
887
There is an important distinction between (a) normal tables, which have probabilities in the body of the table and cut-off values in the margins and (b) tables of t, chi-squared, and F distributions, which have cutoff values in the body of the table and probabilities in the margins.
– BruceET
Jul 29 at 2:59
add a comment |Â
There is an important distinction between (a) normal tables, which have probabilities in the body of the table and cut-off values in the margins and (b) tables of t, chi-squared, and F distributions, which have cutoff values in the body of the table and probabilities in the margins.
– BruceET
Jul 29 at 2:59
There is an important distinction between (a) normal tables, which have probabilities in the body of the table and cut-off values in the margins and (b) tables of t, chi-squared, and F distributions, which have cutoff values in the body of the table and probabilities in the margins.
– BruceET
Jul 29 at 2:59
There is an important distinction between (a) normal tables, which have probabilities in the body of the table and cut-off values in the margins and (b) tables of t, chi-squared, and F distributions, which have cutoff values in the body of the table and probabilities in the margins.
– BruceET
Jul 29 at 2:59
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
This is a Chi Squared test with (n-1) degrees of freedom. $8.58$ is between $.99$ and $.975$ with a probability equal to the area to the left between $.01$ and $.025$. Because this is not less than $.01$, we cannot reject the null hence answer B.
FYI, variance is the "average" squared distance from the mean.
answered Jul 28 at 23:34
Phil H
1,7862311
1
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
1
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
add a comment |Â
up vote
1
down vote
First, intuition:
(1) Value of the sample variance: We can see that $S^2 approx 0.017,$ which is noticeably below the null value $sigma_0^2 = 0.04.$ So it seems we may reject $H_0$ in favor of the left-sided alternative. However, variance estimates based on small samples aren't very precise, so we'll reserve judgment until we see the P-values.
(2) Distribution and value of the test statistic: The test statistic
$Q = frac(n-1)S^2sigma_0^2 = frac20S^2.04 = 8.58.$ But under $H_0,$
$Q sim mathsfChisq(20),$ which has $E(Q) = 20.$ Thus 8.58 is in the lower
tail of the null distribution. Again this raises suspicion we might reject $H_0.$
Second, look closely at the distribution of $mathsfChisq(20)$ for information on the P-value:
Printed table: In a printed chi-squared distribution table,
8.58 is between 8.206 and 9.591 (on row $df = 20)$. The respective headers for the relevant columns show "percentage points"
.99 and .975. This means that 99% of the probability in $mathsfChisq(20)$ is above 8.206 (and 1% below); similarly, 97.5% of the probability is above 9.591
(and 2.5% below). So the P-value is between 0.01 and 0.025.
Beginning students often confuse (i)
"percentage points" in printed tables, which refer to right-tail probabilities, one the one hand, with (ii) the CDF, which refers to left-tail probabilities, on the other hand. The point of this problem may have been to emphasize the difference.
R statistical software: The exact probability below 8.58 can be found using
the chi-squared CDF function pchisq. The result is 0.0127. This is smaller than 0.05, so we would reject $H_0$ at the 5% level. But the P-value $0.0127 > 0.01$
so we cannot reject at the 1% level.
pchisq(8.58, 20)
## 0.01271886
P-values have come into frequent use with the increasing availability of statistical software. Often, you can't find an exact P-value from a printed table--only bracket its value between two numbers in a table (as above).
By contrast, software makes it possible to find exact P-values.
The figure below shows the density function of $mathsfChisq(20)$ along with
the observed value $Q$ of the test statistic. The area under the curve to the left of the vertical broken line is the P-value 0.0127.
answered Jul 29 at 2:24
BruceET
33.1k61440
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is a Chi Squared test with (n-1) degrees of freedom. $8.58$ is between $.99$ and $.975$ with a probability equal to the area to the left between $.01$ and $.025$. Because this is not less than $.01$, we cannot reject the null hence answer B.
FYI, variance is the "average" squared distance from the mean.
answered Jul 28 at 23:34
Phil H
1,7862311
1
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
1
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
add a comment |Â
up vote
1
down vote
This is a Chi Squared test with (n-1) degrees of freedom. $8.58$ is between $.99$ and $.975$ with a probability equal to the area to the left between $.01$ and $.025$. Because this is not less than $.01$, we cannot reject the null hence answer B.
FYI, variance is the "average" squared distance from the mean.
answered Jul 28 at 23:34
Phil H
1,7862311
1
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
1
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is a Chi Squared test with (n-1) degrees of freedom. $8.58$ is between $.99$ and $.975$ with a probability equal to the area to the left between $.01$ and $.025$. Because this is not less than $.01$, we cannot reject the null hence answer B.
FYI, variance is the "average" squared distance from the mean.
answered Jul 28 at 23:34
Phil H
1,7862311
This is a Chi Squared test with (n-1) degrees of freedom. $8.58$ is between $.99$ and $.975$ with a probability equal to the area to the left between $.01$ and $.025$. Because this is not less than $.01$, we cannot reject the null hence answer B.
FYI, variance is the "average" squared distance from the mean.
answered Jul 28 at 23:34
Phil H
1,7862311
edited Jul 29 at 3:18
answered Jul 28 at 23:34
Phil H
1,7862311
answered Jul 28 at 23:34
Phil H
1,7862311
answered Jul 28 at 23:34
Phil H
1,7862311
1,7862311
1
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
1
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
add a comment |Â
1
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
1
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
1
1
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
Why do you think the critical value here is much more than $20$ rather than much less than $20$?
– Henry
Jul 29 at 0:00
1
1
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
@Henry Good point, a smaller $s^2$ would reduce the test statistic yet it would be more likely to be <.04. Do I have this backwards?
– Phil H
Jul 29 at 0:24
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
Frequently, there is confusion between 'percentage point' labels in printed tables, and CDF values. // Successful revision (+1).
– BruceET
Jul 29 at 15:10
add a comment |Â
up vote
1
down vote
First, intuition:
(1) Value of the sample variance: We can see that $S^2 approx 0.017,$ which is noticeably below the null value $sigma_0^2 = 0.04.$ So it seems we may reject $H_0$ in favor of the left-sided alternative. However, variance estimates based on small samples aren't very precise, so we'll reserve judgment until we see the P-values.
(2) Distribution and value of the test statistic: The test statistic
$Q = frac(n-1)S^2sigma_0^2 = frac20S^2.04 = 8.58.$ But under $H_0,$
$Q sim mathsfChisq(20),$ which has $E(Q) = 20.$ Thus 8.58 is in the lower
tail of the null distribution. Again this raises suspicion we might reject $H_0.$
Second, look closely at the distribution of $mathsfChisq(20)$ for information on the P-value:
Printed table: In a printed chi-squared distribution table,
8.58 is between 8.206 and 9.591 (on row $df = 20)$. The respective headers for the relevant columns show "percentage points"
.99 and .975. This means that 99% of the probability in $mathsfChisq(20)$ is above 8.206 (and 1% below); similarly, 97.5% of the probability is above 9.591
(and 2.5% below). So the P-value is between 0.01 and 0.025.
Beginning students often confuse (i)
"percentage points" in printed tables, which refer to right-tail probabilities, one the one hand, with (ii) the CDF, which refers to left-tail probabilities, on the other hand. The point of this problem may have been to emphasize the difference.
R statistical software: The exact probability below 8.58 can be found using
the chi-squared CDF function pchisq. The result is 0.0127. This is smaller than 0.05, so we would reject $H_0$ at the 5% level. But the P-value $0.0127 > 0.01$
so we cannot reject at the 1% level.
pchisq(8.58, 20)
## 0.01271886
P-values have come into frequent use with the increasing availability of statistical software. Often, you can't find an exact P-value from a printed table--only bracket its value between two numbers in a table (as above).
By contrast, software makes it possible to find exact P-values.
The figure below shows the density function of $mathsfChisq(20)$ along with
the observed value $Q$ of the test statistic. The area under the curve to the left of the vertical broken line is the P-value 0.0127.
answered Jul 29 at 2:24
BruceET
33.1k61440
add a comment |Â
up vote
1
down vote
First, intuition:
(1) Value of the sample variance: We can see that $S^2 approx 0.017,$ which is noticeably below the null value $sigma_0^2 = 0.04.$ So it seems we may reject $H_0$ in favor of the left-sided alternative. However, variance estimates based on small samples aren't very precise, so we'll reserve judgment until we see the P-values.
(2) Distribution and value of the test statistic: The test statistic
$Q = frac(n-1)S^2sigma_0^2 = frac20S^2.04 = 8.58.$ But under $H_0,$
$Q sim mathsfChisq(20),$ which has $E(Q) = 20.$ Thus 8.58 is in the lower
tail of the null distribution. Again this raises suspicion we might reject $H_0.$
Second, look closely at the distribution of $mathsfChisq(20)$ for information on the P-value:
Printed table: In a printed chi-squared distribution table,
8.58 is between 8.206 and 9.591 (on row $df = 20)$. The respective headers for the relevant columns show "percentage points"
.99 and .975. This means that 99% of the probability in $mathsfChisq(20)$ is above 8.206 (and 1% below); similarly, 97.5% of the probability is above 9.591
(and 2.5% below). So the P-value is between 0.01 and 0.025.
Beginning students often confuse (i)
"percentage points" in printed tables, which refer to right-tail probabilities, one the one hand, with (ii) the CDF, which refers to left-tail probabilities, on the other hand. The point of this problem may have been to emphasize the difference.
R statistical software: The exact probability below 8.58 can be found using
the chi-squared CDF function pchisq. The result is 0.0127. This is smaller than 0.05, so we would reject $H_0$ at the 5% level. But the P-value $0.0127 > 0.01$
so we cannot reject at the 1% level.
pchisq(8.58, 20)
## 0.01271886
P-values have come into frequent use with the increasing availability of statistical software. Often, you can't find an exact P-value from a printed table--only bracket its value between two numbers in a table (as above).
By contrast, software makes it possible to find exact P-values.
The figure below shows the density function of $mathsfChisq(20)$ along with
the observed value $Q$ of the test statistic. The area under the curve to the left of the vertical broken line is the P-value 0.0127.
answered Jul 29 at 2:24
BruceET
33.1k61440
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First, intuition:
(1) Value of the sample variance: We can see that $S^2 approx 0.017,$ which is noticeably below the null value $sigma_0^2 = 0.04.$ So it seems we may reject $H_0$ in favor of the left-sided alternative. However, variance estimates based on small samples aren't very precise, so we'll reserve judgment until we see the P-values.
(2) Distribution and value of the test statistic: The test statistic
$Q = frac(n-1)S^2sigma_0^2 = frac20S^2.04 = 8.58.$ But under $H_0,$
$Q sim mathsfChisq(20),$ which has $E(Q) = 20.$ Thus 8.58 is in the lower
tail of the null distribution. Again this raises suspicion we might reject $H_0.$
Second, look closely at the distribution of $mathsfChisq(20)$ for information on the P-value:
Printed table: In a printed chi-squared distribution table,
8.58 is between 8.206 and 9.591 (on row $df = 20)$. The respective headers for the relevant columns show "percentage points"
.99 and .975. This means that 99% of the probability in $mathsfChisq(20)$ is above 8.206 (and 1% below); similarly, 97.5% of the probability is above 9.591
(and 2.5% below). So the P-value is between 0.01 and 0.025.
Beginning students often confuse (i)
"percentage points" in printed tables, which refer to right-tail probabilities, one the one hand, with (ii) the CDF, which refers to left-tail probabilities, on the other hand. The point of this problem may have been to emphasize the difference.
R statistical software: The exact probability below 8.58 can be found using
the chi-squared CDF function pchisq. The result is 0.0127. This is smaller than 0.05, so we would reject $H_0$ at the 5% level. But the P-value $0.0127 > 0.01$
so we cannot reject at the 1% level.
pchisq(8.58, 20)
## 0.01271886
P-values have come into frequent use with the increasing availability of statistical software. Often, you can't find an exact P-value from a printed table--only bracket its value between two numbers in a table (as above).
By contrast, software makes it possible to find exact P-values.
The figure below shows the density function of $mathsfChisq(20)$ along with
the observed value $Q$ of the test statistic. The area under the curve to the left of the vertical broken line is the P-value 0.0127.
answered Jul 29 at 2:24
BruceET
33.1k61440
First, intuition:
(1) Value of the sample variance: We can see that $S^2 approx 0.017,$ which is noticeably below the null value $sigma_0^2 = 0.04.$ So it seems we may reject $H_0$ in favor of the left-sided alternative. However, variance estimates based on small samples aren't very precise, so we'll reserve judgment until we see the P-values.
(2) Distribution and value of the test statistic: The test statistic
$Q = frac(n-1)S^2sigma_0^2 = frac20S^2.04 = 8.58.$ But under $H_0,$
$Q sim mathsfChisq(20),$ which has $E(Q) = 20.$ Thus 8.58 is in the lower
tail of the null distribution. Again this raises suspicion we might reject $H_0.$
Second, look closely at the distribution of $mathsfChisq(20)$ for information on the P-value:
Printed table: In a printed chi-squared distribution table,
8.58 is between 8.206 and 9.591 (on row $df = 20)$. The respective headers for the relevant columns show "percentage points"
.99 and .975. This means that 99% of the probability in $mathsfChisq(20)$ is above 8.206 (and 1% below); similarly, 97.5% of the probability is above 9.591
(and 2.5% below). So the P-value is between 0.01 and 0.025.
Beginning students often confuse (i)
"percentage points" in printed tables, which refer to right-tail probabilities, one the one hand, with (ii) the CDF, which refers to left-tail probabilities, on the other hand. The point of this problem may have been to emphasize the difference.
R statistical software: The exact probability below 8.58 can be found using
the chi-squared CDF function pchisq. The result is 0.0127. This is smaller than 0.05, so we would reject $H_0$ at the 5% level. But the P-value $0.0127 > 0.01$
so we cannot reject at the 1% level.
pchisq(8.58, 20)
## 0.01271886
P-values have come into frequent use with the increasing availability of statistical software. Often, you can't find an exact P-value from a printed table--only bracket its value between two numbers in a table (as above).
By contrast, software makes it possible to find exact P-values.
The figure below shows the density function of $mathsfChisq(20)$ along with
the observed value $Q$ of the test statistic. The area under the curve to the left of the vertical broken line is the P-value 0.0127.
answered Jul 29 at 2:24
BruceET
33.1k61440
edited Jul 29 at 15:08
answered Jul 29 at 2:24
BruceET
33.1k61440
answered Jul 29 at 2:24
BruceET
33.1k61440
answered Jul 29 at 2:24
BruceET
33.1k61440
33.1k61440
add a comment |Â
add a comment |Â
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There is an important distinction between (a) normal tables, which have probabilities in the body of the table and cut-off values in the margins and (b) tables of t, chi-squared, and F distributions, which have cutoff values in the body of the table and probabilities in the margins.
– BruceET
Jul 29 at 2:59