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Statistics and Confidence Intervals

Construct a Confidence Interval of $95%$Confidence level of random sample from continuous distributionConfidence interval - No sampleInterval estimate to infer the population mean with a 95% confidence levelConfidence IntervalsConfidence Intervals Calculation and InterpretationConfidence intervals and significance testsHow to determine a confidence interval without knowing mean and varianceReducing the confidence interval with T distribitionConfidence Intervals using Pivotal Quantities

0

$begingroup$

Given the following set of values:

10,11,14,95,73,30,29,9,97,94,70

How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10

Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:

$x-z_fraca2 fracsigma^2sqrt(n)$ and $x+z_fraca2 fracsigma^2sqrt(n)$ where x denotes the mean. Here the significance level (a) is 0.01.

statistics probability-distributions normal-distribution statistical-inference

share|cite|improve this question

edited Mar 21 at 18:11

topologicalmagician

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts? Are you familiar with any formulas that concern confidence intervals? What computations can you do?
  $endgroup$
  – Brian
  Mar 21 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I know the formula for the confidence interval, but i'm not sure how to find the z-score, I know that I can use R, but i'm not sure what to write in R.
  $endgroup$
  – topologicalmagician
  Mar 21 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what the "99%" in the confidence interval means? Can you relate this to the Normal distribution in some way? Do you know how z-scores relate to the Normal distribution?
  $endgroup$
  – Brian
  Mar 21 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Brian Yes, a CI is the 100(1-a)% interval in which the true parameter lies in, a is the significance level. The formula for the CI is : $x+z_fraca2 fracsigma^2sqrt(n)$ and : $x-z_fraca2 fracsigma^2sqrt(n)$ where x denotes the mean
  $endgroup$
  – topologicalmagician
  Mar 21 at 18:04
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Given the following set of values:

10,11,14,95,73,30,29,9,97,94,70

How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10

Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:

$x-z_fraca2 fracsigma^2sqrt(n)$ and $x+z_fraca2 fracsigma^2sqrt(n)$ where x denotes the mean. Here the significance level (a) is 0.01.

statistics probability-distributions normal-distribution statistical-inference

share|cite|improve this question

edited Mar 21 at 18:11

topologicalmagician

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts? Are you familiar with any formulas that concern confidence intervals? What computations can you do?
  $endgroup$
  – Brian
  Mar 21 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I know the formula for the confidence interval, but i'm not sure how to find the z-score, I know that I can use R, but i'm not sure what to write in R.
  $endgroup$
  – topologicalmagician
  Mar 21 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what the "99%" in the confidence interval means? Can you relate this to the Normal distribution in some way? Do you know how z-scores relate to the Normal distribution?
  $endgroup$
  – Brian
  Mar 21 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Brian Yes, a CI is the 100(1-a)% interval in which the true parameter lies in, a is the significance level. The formula for the CI is : $x+z_fraca2 fracsigma^2sqrt(n)$ and : $x-z_fraca2 fracsigma^2sqrt(n)$ where x denotes the mean
  $endgroup$
  – topologicalmagician
  Mar 21 at 18:04
  
  

  
  
  
  

add a comment | 

$begingroup$

Given the following set of values:

10,11,14,95,73,30,29,9,97,94,70

How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10

Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:

$x-z_fraca2 fracsigma^2sqrt(n)$ and $x+z_fraca2 fracsigma^2sqrt(n)$ where x denotes the mean. Here the significance level (a) is 0.01.

statistics probability-distributions normal-distribution statistical-inference

share|cite|improve this question

edited Mar 21 at 18:11

topologicalmagician

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

$endgroup$

share|cite|improve this question

edited Mar 21 at 18:11

topologicalmagician

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

share|cite|improve this question

edited Mar 21 at 18:11

topologicalmagician

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

asked Mar 21 at 17:50

topologicalmagiciantopologicalmagician

1199

1 Answer
1

active

oldest

votes

1

$begingroup$

It is likely that you are asked to find a confidence interval for the (unknown) population mean, not the sample mean. The sample mean is not a parameter of interest, it can be calculated from the sample.

A conventional setup for the problem is that you have a sample $mathbf X=(X_1,X_2,ldots,X_n)$ of size $n=11$ from a $N(mu,sigma_0^2)$ population (by assumption) with $sigma_0^2=10$. You have to find a confidence interval for the mean $mu$.

A suitable pivotal quantity here is $$Q(mathbf X,mu)=fracsqrtn(overline X-mu)sigma_0sim N(0,1)$$

, where $overline X=frac1nsumlimits_i=1^n X_i$ is the sample mean.

So if $z_alpha/2$ be such that $P(Z>z_alpha/2)=alpha/2$ where $Zsim N(0,1)$, you have $$P_mu(-z_alpha/2le Qle z_alpha/2)=1-alphaquad,forall,mu$$

You have to use the above to arrive at the form $$P_mu(c_1,le mule,, c_2)=1-alphaquad,forall,mu$$

A $100(1-alpha)%$ confidence interval for $mu$ is then $[c_1,c_2]$.

share|cite|improve this answer

edited Mar 21 at 19:07

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440

$endgroup$

add a comment | 

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1

$begingroup$

It is likely that you are asked to find a confidence interval for the (unknown) population mean, not the sample mean. The sample mean is not a parameter of interest, it can be calculated from the sample.

A conventional setup for the problem is that you have a sample $mathbf X=(X_1,X_2,ldots,X_n)$ of size $n=11$ from a $N(mu,sigma_0^2)$ population (by assumption) with $sigma_0^2=10$. You have to find a confidence interval for the mean $mu$.

A suitable pivotal quantity here is $$Q(mathbf X,mu)=fracsqrtn(overline X-mu)sigma_0sim N(0,1)$$

, where $overline X=frac1nsumlimits_i=1^n X_i$ is the sample mean.

So if $z_alpha/2$ be such that $P(Z>z_alpha/2)=alpha/2$ where $Zsim N(0,1)$, you have $$P_mu(-z_alpha/2le Qle z_alpha/2)=1-alphaquad,forall,mu$$

You have to use the above to arrive at the form $$P_mu(c_1,le mule,, c_2)=1-alphaquad,forall,mu$$

A $100(1-alpha)%$ confidence interval for $mu$ is then $[c_1,c_2]$.

share|cite|improve this answer

edited Mar 21 at 19:07

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440

$endgroup$

add a comment | 

1

$begingroup$

It is likely that you are asked to find a confidence interval for the (unknown) population mean, not the sample mean. The sample mean is not a parameter of interest, it can be calculated from the sample.

A conventional setup for the problem is that you have a sample $mathbf X=(X_1,X_2,ldots,X_n)$ of size $n=11$ from a $N(mu,sigma_0^2)$ population (by assumption) with $sigma_0^2=10$. You have to find a confidence interval for the mean $mu$.

A suitable pivotal quantity here is $$Q(mathbf X,mu)=fracsqrtn(overline X-mu)sigma_0sim N(0,1)$$

, where $overline X=frac1nsumlimits_i=1^n X_i$ is the sample mean.

So if $z_alpha/2$ be such that $P(Z>z_alpha/2)=alpha/2$ where $Zsim N(0,1)$, you have $$P_mu(-z_alpha/2le Qle z_alpha/2)=1-alphaquad,forall,mu$$

You have to use the above to arrive at the form $$P_mu(c_1,le mule,, c_2)=1-alphaquad,forall,mu$$

A $100(1-alpha)%$ confidence interval for $mu$ is then $[c_1,c_2]$.

share|cite|improve this answer

edited Mar 21 at 19:07

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440

$endgroup$

add a comment | 

$begingroup$

It is likely that you are asked to find a confidence interval for the (unknown) population mean, not the sample mean. The sample mean is not a parameter of interest, it can be calculated from the sample.

A conventional setup for the problem is that you have a sample $mathbf X=(X_1,X_2,ldots,X_n)$ of size $n=11$ from a $N(mu,sigma_0^2)$ population (by assumption) with $sigma_0^2=10$. You have to find a confidence interval for the mean $mu$.

A suitable pivotal quantity here is $$Q(mathbf X,mu)=fracsqrtn(overline X-mu)sigma_0sim N(0,1)$$

, where $overline X=frac1nsumlimits_i=1^n X_i$ is the sample mean.

So if $z_alpha/2$ be such that $P(Z>z_alpha/2)=alpha/2$ where $Zsim N(0,1)$, you have $$P_mu(-z_alpha/2le Qle z_alpha/2)=1-alphaquad,forall,mu$$

You have to use the above to arrive at the form $$P_mu(c_1,le mule,, c_2)=1-alphaquad,forall,mu$$

A $100(1-alpha)%$ confidence interval for $mu$ is then $[c_1,c_2]$.

share|cite|improve this answer

edited Mar 21 at 19:07

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440

$endgroup$

share|cite|improve this answer

edited Mar 21 at 19:07

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440

answered Mar 21 at 18:56

StubbornAtomStubbornAtom

6,29831440