What are the values of a student's distribution (k, µ and σ) when x1 = 10, x2 = 10 and x3 = 13 and then 10?...
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What are the values of a student's distribution (k, µ and σ) when x1 = 10, x2 = 10 and x3 = 13 and then 10?
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$begingroup$
I have a sample that contains only 3 values that are
x1 = 10
x2 = 10
x3 = 13
Remark: I know that x3 value is not what is written in title
I will calculate the probability that a new value is greater than the limit.
I think that I can use T-Student distribution to compute this probability.
My problem is to find the parameter k, µ and σ to know the T-Student distribution that I can use to compute this probability.
The sample size n = 3.
The mean can be computed using following formula
$$
overline{x} = frac{1}{n} sum_{i=1}^n x_i
$$
His value is 33/3 = 11
The degree of freedom k = n-1 = 2
The variance can be estimated using following formula.
$$
S = tfrac{1}{n-1}sum_{i=1}^n (x_i - overline{x}) ^2
$$
His value is (1+1+2*2)/2 = 3.
If value of x3 is not more 13 BUT 10, what is variance value ?
0.0 !
In real world, where precision of measure is 0.5, how can I resolve this issue ?
This question is very important for me, because in first case I can use T-Student distribution to compute a confidence interval, but when variance is zero it is not more possible !
I'm interesting to know how engineers resolve this issue in real world when doing new other tests are not possible ?
What happens ? This question is too complex ! What is your experiment when this situation happen ?
probability-distributions
asked Feb 15 at 7:39
schlebeschlebe
1036
$endgroup$
add a comment |
$begingroup$
I have a sample that contains only 3 values that are
x1 = 10
x2 = 10
x3 = 13
Remark: I know that x3 value is not what is written in title
I will calculate the probability that a new value is greater than the limit.
I think that I can use T-Student distribution to compute this probability.
My problem is to find the parameter k, µ and σ to know the T-Student distribution that I can use to compute this probability.
The sample size n = 3.
The mean can be computed using following formula
$$
overline{x} = frac{1}{n} sum_{i=1}^n x_i
$$
His value is 33/3 = 11
The degree of freedom k = n-1 = 2
The variance can be estimated using following formula.
$$
S = tfrac{1}{n-1}sum_{i=1}^n (x_i - overline{x}) ^2
$$
His value is (1+1+2*2)/2 = 3.
If value of x3 is not more 13 BUT 10, what is variance value ?
0.0 !
In real world, where precision of measure is 0.5, how can I resolve this issue ?
This question is very important for me, because in first case I can use T-Student distribution to compute a confidence interval, but when variance is zero it is not more possible !
I'm interesting to know how engineers resolve this issue in real world when doing new other tests are not possible ?
What happens ? This question is too complex ! What is your experiment when this situation happen ?
probability-distributions
asked Feb 15 at 7:39
schlebeschlebe
1036
$endgroup$
add a comment |
$begingroup$
I have a sample that contains only 3 values that are
x1 = 10
x2 = 10
x3 = 13
Remark: I know that x3 value is not what is written in title
I will calculate the probability that a new value is greater than the limit.
I think that I can use T-Student distribution to compute this probability.
My problem is to find the parameter k, µ and σ to know the T-Student distribution that I can use to compute this probability.
The sample size n = 3.
The mean can be computed using following formula
$$
overline{x} = frac{1}{n} sum_{i=1}^n x_i
$$
His value is 33/3 = 11
The degree of freedom k = n-1 = 2
The variance can be estimated using following formula.
$$
S = tfrac{1}{n-1}sum_{i=1}^n (x_i - overline{x}) ^2
$$
His value is (1+1+2*2)/2 = 3.
If value of x3 is not more 13 BUT 10, what is variance value ?
0.0 !
In real world, where precision of measure is 0.5, how can I resolve this issue ?
This question is very important for me, because in first case I can use T-Student distribution to compute a confidence interval, but when variance is zero it is not more possible !
I'm interesting to know how engineers resolve this issue in real world when doing new other tests are not possible ?
What happens ? This question is too complex ! What is your experiment when this situation happen ?
probability-distributions
asked Feb 15 at 7:39
schlebeschlebe
1036
$endgroup$
I have a sample that contains only 3 values that are
x1 = 10
x2 = 10
x3 = 13
Remark: I know that x3 value is not what is written in title
I will calculate the probability that a new value is greater than the limit.
I think that I can use T-Student distribution to compute this probability.
My problem is to find the parameter k, µ and σ to know the T-Student distribution that I can use to compute this probability.
The sample size n = 3.
The mean can be computed using following formula
$$
overline{x} = frac{1}{n} sum_{i=1}^n x_i
$$
His value is 33/3 = 11
The degree of freedom k = n-1 = 2
The variance can be estimated using following formula.
$$
S = tfrac{1}{n-1}sum_{i=1}^n (x_i - overline{x}) ^2
$$
His value is (1+1+2*2)/2 = 3.
If value of x3 is not more 13 BUT 10, what is variance value ?
0.0 !
In real world, where precision of measure is 0.5, how can I resolve this issue ?
This question is very important for me, because in first case I can use T-Student distribution to compute a confidence interval, but when variance is zero it is not more possible !
I'm interesting to know how engineers resolve this issue in real world when doing new other tests are not possible ?
What happens ? This question is too complex ! What is your experiment when this situation happen ?
probability-distributions
probability-distributions
asked Feb 15 at 7:39
schlebeschlebe
1036
asked Feb 15 at 7:39
schlebeschlebe
1036
edited Mar 16 at 5:49
schlebe
asked Feb 15 at 7:39
schlebeschlebe
1036
asked Feb 15 at 7:39
schlebeschlebe
1036
asked Feb 15 at 7:39
schlebeschlebe
1036
1036
add a comment |
add a comment |
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