Mixing
Normal Distribution Or Student Distribution?
$begingroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value. For statistical purposes, you decide to throw $10$ fifty-cent coins in vending machine A. This results in a sample mean of $7.49$ $g$ and a sample variance of $0.011$ $g^2$. You may assume a normal distribution as model distribution for the measurements.
Construct a $95%$ confidence interval for the mean weight $μ_A$ of a fity-cent coin thrown in machine $A$.
In this exercise, you'll use the student distribution or the standard normal distribution to construct the confidence interval? I know that they say 'You may assume a normal distribution as model distribution for the measurements' but they also give you the sample variance (that is used with the student distribution) and not the variance. Furthermore, $n=10$ is quite small, so, in my opinion, I should use the student distribution, it's correct?
statistics
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value. For statistical purposes, you decide to throw $10$ fifty-cent coins in vending machine A. This results in a sample mean of $7.49$ $g$ and a sample variance of $0.011$ $g^2$. You may assume a normal distribution as model distribution for the measurements.
Construct a $95%$ confidence interval for the mean weight $μ_A$ of a fity-cent coin thrown in machine $A$.
In this exercise, you'll use the student distribution or the standard normal distribution to construct the confidence interval? I know that they say 'You may assume a normal distribution as model distribution for the measurements' but they also give you the sample variance (that is used with the student distribution) and not the variance. Furthermore, $n=10$ is quite small, so, in my opinion, I should use the student distribution, it's correct?
statistics
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
$endgroup$
1
$begingroup$
Is the question which distribution to use part of the exercise?
$endgroup$
– I like Serena
Jan 22 at 22:17
1
$begingroup$
No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value.
$endgroup$
– Mark Jacon
Jan 22 at 22:35
1
$begingroup$
The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution.
$endgroup$
– I like Serena
Jan 23 at 7:31
add a comment |
$begingroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value. For statistical purposes, you decide to throw $10$ fifty-cent coins in vending machine A. This results in a sample mean of $7.49$ $g$ and a sample variance of $0.011$ $g^2$. You may assume a normal distribution as model distribution for the measurements.
Construct a $95%$ confidence interval for the mean weight $μ_A$ of a fity-cent coin thrown in machine $A$.
In this exercise, you'll use the student distribution or the standard normal distribution to construct the confidence interval? I know that they say 'You may assume a normal distribution as model distribution for the measurements' but they also give you the sample variance (that is used with the student distribution) and not the variance. Furthermore, $n=10$ is quite small, so, in my opinion, I should use the student distribution, it's correct?
statistics
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
$endgroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value. For statistical purposes, you decide to throw $10$ fifty-cent coins in vending machine A. This results in a sample mean of $7.49$ $g$ and a sample variance of $0.011$ $g^2$. You may assume a normal distribution as model distribution for the measurements.
Construct a $95%$ confidence interval for the mean weight $μ_A$ of a fity-cent coin thrown in machine $A$.
In this exercise, you'll use the student distribution or the standard normal distribution to construct the confidence interval? I know that they say 'You may assume a normal distribution as model distribution for the measurements' but they also give you the sample variance (that is used with the student distribution) and not the variance. Furthermore, $n=10$ is quite small, so, in my opinion, I should use the student distribution, it's correct?
statistics
statistics
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
asked Jan 22 at 22:02
Mark JaconMark Jacon
1127
1127
1
$begingroup$
Is the question which distribution to use part of the exercise?
$endgroup$
– I like Serena
Jan 22 at 22:17
1
$begingroup$
No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value.
$endgroup$
– Mark Jacon
Jan 22 at 22:35
1
$begingroup$
The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution.
$endgroup$
– I like Serena
Jan 23 at 7:31
add a comment |
1
$begingroup$
Is the question which distribution to use part of the exercise?
$endgroup$
– I like Serena
Jan 22 at 22:17
1
$begingroup$
No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value.
$endgroup$
– Mark Jacon
Jan 22 at 22:35
1
$begingroup$
The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution.
$endgroup$
– I like Serena
Jan 23 at 7:31
1
1
$begingroup$
Is the question which distribution to use part of the exercise?
$endgroup$
– I like Serena
Jan 22 at 22:17
$begingroup$
Is the question which distribution to use part of the exercise?
$endgroup$
– I like Serena
Jan 22 at 22:17
1
1
$begingroup$
No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value.
$endgroup$
– Mark Jacon
Jan 22 at 22:35
$begingroup$
No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value.
$endgroup$
– Mark Jacon
Jan 22 at 22:35
1
1
$begingroup$
The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution.
$endgroup$
– I like Serena
Jan 23 at 7:31
$begingroup$
The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution.
$endgroup$
– I like Serena
Jan 23 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are correct that you need to use Student's t distribution with $n - 1$ $= 10 - 1 = 9$ degrees of freedom. That's because the population variance $sigma^2$ is estimated by the sample variance $S^2 = 0.011.$ (Strictly speaking, the sample size is not relevant to the decision to use the t distribution instead of the normal distribution.)
The formula to get a 95% confidence interval for the population mean
$mu$ is $bar X pm t^*sqrt{S^2/n},$ where $t^*$ cuts area 2.5% from the upper tail of $mathsf{T}(text{df}=9).$ Using, R statistical software,
I get the value for $t^*$ shown below. If your course uses printed
tables, you should look at the appropriate table now to find essentially the same number (typically, with fewer decimal places).
qt(.975, 9)
## 2.262157
Notes: For a 95% CI and a sample of size $n = 35,$ software gives $t^* =2.0322,$
as shown below. If you were given the exact value of the population variance $sigma^2,$ then you could use the formula $bar X pm 1.96sqrt{sigma^2/n},$ where the value $z^* = 1.960$ cuts 2.5% of the area from
the upper tail of a standard normal distribution.
qt(.975, 34)
## 2.032245
qnorm(.975)
## 1.959964
For $n > 30$ and 95%
confidence intervals, both $t^*$ and $z^* =1.960$ round to $2.0.$ That's
why some of the less-fussy textbooks say you can use $z^*$ instead of $t^*$ for $n > 30.$ However, this
"rule of 30" works only for 95% confidence intervals.
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
$endgroup$
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
add a comment |
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$begingroup$
You are correct that you need to use Student's t distribution with $n - 1$ $= 10 - 1 = 9$ degrees of freedom. That's because the population variance $sigma^2$ is estimated by the sample variance $S^2 = 0.011.$ (Strictly speaking, the sample size is not relevant to the decision to use the t distribution instead of the normal distribution.)
The formula to get a 95% confidence interval for the population mean
$mu$ is $bar X pm t^*sqrt{S^2/n},$ where $t^*$ cuts area 2.5% from the upper tail of $mathsf{T}(text{df}=9).$ Using, R statistical software,
I get the value for $t^*$ shown below. If your course uses printed
tables, you should look at the appropriate table now to find essentially the same number (typically, with fewer decimal places).
qt(.975, 9)
## 2.262157
Notes: For a 95% CI and a sample of size $n = 35,$ software gives $t^* =2.0322,$
as shown below. If you were given the exact value of the population variance $sigma^2,$ then you could use the formula $bar X pm 1.96sqrt{sigma^2/n},$ where the value $z^* = 1.960$ cuts 2.5% of the area from
the upper tail of a standard normal distribution.
qt(.975, 34)
## 2.032245
qnorm(.975)
## 1.959964
For $n > 30$ and 95%
confidence intervals, both $t^*$ and $z^* =1.960$ round to $2.0.$ That's
why some of the less-fussy textbooks say you can use $z^*$ instead of $t^*$ for $n > 30.$ However, this
"rule of 30" works only for 95% confidence intervals.
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
$endgroup$
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
add a comment |
$begingroup$
You are correct that you need to use Student's t distribution with $n - 1$ $= 10 - 1 = 9$ degrees of freedom. That's because the population variance $sigma^2$ is estimated by the sample variance $S^2 = 0.011.$ (Strictly speaking, the sample size is not relevant to the decision to use the t distribution instead of the normal distribution.)
The formula to get a 95% confidence interval for the population mean
$mu$ is $bar X pm t^*sqrt{S^2/n},$ where $t^*$ cuts area 2.5% from the upper tail of $mathsf{T}(text{df}=9).$ Using, R statistical software,
I get the value for $t^*$ shown below. If your course uses printed
tables, you should look at the appropriate table now to find essentially the same number (typically, with fewer decimal places).
qt(.975, 9)
## 2.262157
Notes: For a 95% CI and a sample of size $n = 35,$ software gives $t^* =2.0322,$
as shown below. If you were given the exact value of the population variance $sigma^2,$ then you could use the formula $bar X pm 1.96sqrt{sigma^2/n},$ where the value $z^* = 1.960$ cuts 2.5% of the area from
the upper tail of a standard normal distribution.
qt(.975, 34)
## 2.032245
qnorm(.975)
## 1.959964
For $n > 30$ and 95%
confidence intervals, both $t^*$ and $z^* =1.960$ round to $2.0.$ That's
why some of the less-fussy textbooks say you can use $z^*$ instead of $t^*$ for $n > 30.$ However, this
"rule of 30" works only for 95% confidence intervals.
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
$endgroup$
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
add a comment |
$begingroup$
You are correct that you need to use Student's t distribution with $n - 1$ $= 10 - 1 = 9$ degrees of freedom. That's because the population variance $sigma^2$ is estimated by the sample variance $S^2 = 0.011.$ (Strictly speaking, the sample size is not relevant to the decision to use the t distribution instead of the normal distribution.)
The formula to get a 95% confidence interval for the population mean
$mu$ is $bar X pm t^*sqrt{S^2/n},$ where $t^*$ cuts area 2.5% from the upper tail of $mathsf{T}(text{df}=9).$ Using, R statistical software,
I get the value for $t^*$ shown below. If your course uses printed
tables, you should look at the appropriate table now to find essentially the same number (typically, with fewer decimal places).
qt(.975, 9)
## 2.262157
Notes: For a 95% CI and a sample of size $n = 35,$ software gives $t^* =2.0322,$
as shown below. If you were given the exact value of the population variance $sigma^2,$ then you could use the formula $bar X pm 1.96sqrt{sigma^2/n},$ where the value $z^* = 1.960$ cuts 2.5% of the area from
the upper tail of a standard normal distribution.
qt(.975, 34)
## 2.032245
qnorm(.975)
## 1.959964
For $n > 30$ and 95%
confidence intervals, both $t^*$ and $z^* =1.960$ round to $2.0.$ That's
why some of the less-fussy textbooks say you can use $z^*$ instead of $t^*$ for $n > 30.$ However, this
"rule of 30" works only for 95% confidence intervals.
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
$endgroup$
You are correct that you need to use Student's t distribution with $n - 1$ $= 10 - 1 = 9$ degrees of freedom. That's because the population variance $sigma^2$ is estimated by the sample variance $S^2 = 0.011.$ (Strictly speaking, the sample size is not relevant to the decision to use the t distribution instead of the normal distribution.)
The formula to get a 95% confidence interval for the population mean
$mu$ is $bar X pm t^*sqrt{S^2/n},$ where $t^*$ cuts area 2.5% from the upper tail of $mathsf{T}(text{df}=9).$ Using, R statistical software,
I get the value for $t^*$ shown below. If your course uses printed
tables, you should look at the appropriate table now to find essentially the same number (typically, with fewer decimal places).
qt(.975, 9)
## 2.262157
Notes: For a 95% CI and a sample of size $n = 35,$ software gives $t^* =2.0322,$
as shown below. If you were given the exact value of the population variance $sigma^2,$ then you could use the formula $bar X pm 1.96sqrt{sigma^2/n},$ where the value $z^* = 1.960$ cuts 2.5% of the area from
the upper tail of a standard normal distribution.
qt(.975, 34)
## 2.032245
qnorm(.975)
## 1.959964
For $n > 30$ and 95%
confidence intervals, both $t^*$ and $z^* =1.960$ round to $2.0.$ That's
why some of the less-fussy textbooks say you can use $z^*$ instead of $t^*$ for $n > 30.$ However, this
"rule of 30" works only for 95% confidence intervals.
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
edited Jan 23 at 1:08
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
answered Jan 23 at 1:01
BruceETBruceET
35.9k71540
35.9k71540
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
add a comment |
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
$begingroup$
thanks a lot, now it's clear!
$endgroup$
– Mark Jacon
Jan 23 at 17:19
add a comment |
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1
$begingroup$
Is the question which distribution to use part of the exercise?
$endgroup$
– I like Serena
Jan 22 at 22:17
1
$begingroup$
No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value.
$endgroup$
– Mark Jacon
Jan 22 at 22:35
1
$begingroup$
The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution.
$endgroup$
– I like Serena
Jan 23 at 7:31