Confidence Interval for Mean $\mu$: Let $X_1, X_2, \dots, X_n$

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SUMMARY

The discussion centers on establishing that the interval $$\left( \overline{X}-z_{k=a} \frac{\sigma}{\sqrt{n}}, \overline{X}+z_{(1-k)=a} \frac{\sigma}{\sqrt{n}}\right)$$ serves as a 100(1-a)% confidence interval for the mean $\mu$ of a random sample $X_1, X_2, \dots, X_n$. The parameters involved include the sample mean $\overline{X}$, the standard deviation $\sigma$, and the sample size $n$. The discussion emphasizes the significance of the values $k$ and $z_{k=a}$ in determining the bounds of the confidence interval.

PREREQUISITES
  • Understanding of random sampling and its properties
  • Knowledge of confidence intervals and their construction
  • Familiarity with statistical notation, including $\overline{X}$, $\sigma$, and $n$
  • Basic comprehension of the standard normal distribution and $z$-scores
NEXT STEPS
  • Study the derivation of confidence intervals for the mean using the Central Limit Theorem
  • Learn about the properties of the standard normal distribution and how to find $z$-scores
  • Explore the implications of varying the parameter $a$ on the confidence interval width
  • Investigate the differences between confidence intervals for means with known versus unknown variance
USEFUL FOR

Statisticians, data analysts, and students in statistics who are looking to deepen their understanding of confidence intervals and their applications in inferential statistics.

evinda
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Hello! (Wave)

Let $X_1, X_2, \dots, X_n$ a random sample with $E(X_i)=\mu$, $Var(X_i)=\sigma^2 \forall i$. For $0<a<0.5$:

show that for any $k \in [0,1]$, the interval
$$\left( \overline{X}-z_{k=a} \frac{\sigma}{\sqrt{n}}, \overline{X}+z_{(1-k)=a} \frac{\sigma}{\sqrt{n}}\right)$$
is a 100(1-a)% confidence interval for the mean $\mu$.

How could we do this? (Thinking)
 
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evinda said:
Hello! (Wave)

Let $X_1, X_2, \dots, X_n$ a random sample with $E(X_i)=\mu$, $Var(X_i)=\sigma^2 \forall i$. For $0<a<0.5$:

show that for any $k \in [0,1]$, the interval
$$\left( \overline{X}-z_{k=a} \frac{\sigma}{\sqrt{n}}, \overline{X}+z_{(1-k)=a} \frac{\sigma}{\sqrt{n}}\right)$$
is a 100(1-a)% confidence interval for the mean $\mu$.

How could we do this? (Thinking)

Hey! (Smile)

What is $k$? And what is $z_{k=a}$? (Wondering)
 

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