Solve Sampling Problems: 95% Interval for Mean of X

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Homework Help Overview

The discussion revolves around calculating a 95% confidence interval for the mean of a normally distributed variable X, given its mean and variance. The original poster presents a calculation attempt and seeks clarification on the correct approach to derive the confidence interval.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's calculation method, questioning the use of the variance in the interval formula. There is a focus on ensuring the interval is symmetric about the mean. Some participants suggest modifications to the formula to achieve this symmetry.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's approach and suggesting corrections. There is a recognition of the need for clarity regarding the problem statement and the proper interpretation of the confidence interval.

Contextual Notes

Participants note that the original problem statement may contain contradictions, particularly regarding the interpretation of the mean and the sample mean. There is also mention of the need for unbiased estimates of mean and variance based on provided data.

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Homework Statement



If X is distributed normally with mean = 7 and the variance of X is 4 , calculate a 95% interval for mean of X . size interval is 10

Homework Equations


The Attempt at a Solution



here's my working :
( 4- 1.960 x (surd(4/10)) , 4+1.960 x (surd(4/10)) )

my ans is incorrect . the correct ans is (6.24, 7.59)
trGFvqC.png
formula which i used

[/B]
 
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I assume you mean a confidence interval for the mean of a sample, sample size 10.
How do you get 4-... and 4+...? Think about that again.
But I don't see how the given answer can be right either. It should be symmetric about the mean, no?
 
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somecelxis said:

Homework Statement



If X is distributed normally with mean = 7 and the variance of X is 4 , calculate a 95% interval for mean of X . size interval is 10

Homework Equations


The Attempt at a Solution



here's my working :
( 4- 1.960 x (surd(4/10)) , 4+1.960 x (surd(4/10)) )

my ans is incorrect . the correct ans is (6.24, 7.59)
trGFvqC.png
formula which i used
[/B]

Your problem statement is contradictory as you have written it, but it can be fixed: you want a 95% probability interval for the SAMPLE mean (not the mean, which was already given as 7 exactly). So, you want an interval ##[a,b]## with
\Pr \left( \bar{X} \in [a,b] \right) = 0.95
In principle, the interval ##[a,b]## can have any location, but in practice it is almost always chosen to be symmetric about the mean true mean. Your formula does not reflect that. Do you see how you can modify it to satisfy the symmetry requirement?

Note: maybe you are confusing your problem with one in which the true mean ##\mu## is unknown but must also be estimated from the data itself.
 
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haruspex said:
I assume you mean a confidence interval for the mean of a sample, sample size 10.
How do you get 4-... and 4+...? Think about that again.
But I don't see how the given answer can be right either. It should be symmetric about the mean, no?
( 7- 1.960 x (surd(4/10)) , 7+1.960 x (surd(4/10)) )
 
Ray Vickson said:
Your problem statement is contradictory as you have written it, but it can be fixed: you want a 95% probability interval for the SAMPLE mean (not the mean, which was already given as 7 exactly). So, you want an interval ##[a,b]## with
\Pr \left( \bar{X} \in [a,b] \right) = 0.95
In principle, the interval ##[a,b]## can have any location, but in practice it is almost always chosen to be symmetric about the mean true mean. Your formula does not reflect that. Do you see how you can modify it to satisfy the symmetry requirement?

Note: maybe you are confusing your problem with one in which the true mean ##\mu## is unknown but must also be estimated from the data itself.
please allow me to post the whole question here...
A government wants to study X , the time taken by an employee are chosen and the information obtained is summarised as sum of X= 70 , sum of X^2 = 522 ..
(i) Find unbiased estimate for mean and variance X
(2) If X is distributed normally and the variance of X is 4 . Calculate a 95% confidence interval for the mean of X .
the attached photo is the whole question and my attempt of the solution...
 

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somecelxis said:
please allow me to post the whole question here...
A government wants to study X , the time taken by an employee are chosen and the information obtained is summarised as sum of X= 70 , sum of X^2 = 522 ..
(i) Find unbiased estimate for mean and variance X
(2) If X is distributed normally and the variance of X is 4 . Calculate a 95% confidence interval for the mean of X .
the attached photo is the whole question and my attempt of the solution...

Please do not post thumbnails; they are unreadable on some media. Anyway, why bother posting a thumbnail of the problem, as you seem to have written out the basics of the problem already? Now, just type out the solution.
 
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Ray Vickson said:
Please do not post thumbnails; they are unreadable on some media. Anyway, why bother posting a thumbnail of the problem, as you seem to have written out the basics of the problem already? Now, just type out the solution.
i don't know how to use LATEX.
estimated mean = 70/10 = 7
estimated variance = 1/(10-1) x ( 522 - (70x70)/10 ) = 32/9
symmetrical interval = 7- 1.960 surd (4/10) , 7+ 1.960 surd (4/10)
 
somecelxis said:
i don't know how to use LATEX.
estimated mean = 70/10 = 7
estimated variance = 1/(10-1) x ( 522 - (70x70)/10 ) = 32/9
symmetrical interval = 7- 1.960 surd (4/10) , 7+ 1.960 surd (4/10)

No need to use LaTeX; what you typed here is perfectly legible, although it is more usual to say sqrt(x) rather than surd(x).

BTW: your answer is correct.
 

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