- #1
A_B
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Homework Statement
in 225 out of 2700 schools, the number of registrations were recorded. The average was 3700, and the standard deviation 6000. Give the 95% confidence interval for the total number of registrations in all of the 2700 schools.
Homework Equations
The Attempt at a Solution
I first found the distribution of the number of registrations in one school.
[tex] T_I=\frac{\bar{I}-\mu}{S/ \sqrt{n}} \sim t_{n-1} [/tex]
Which is distributed as student's t distribution.
I then found the expected value and the variation of [itex]\bar{I}[/itex].
[tex]E(\bar{I})=\mu[/tex]
[tex]
\begin{align*}
Var(T_I)=\frac{n-1}{n-3} &= Var(\frac{\sqrt(n)}{S} \bar{I} - \frac{\mu\sqrt(n)}{S} \\
\frac{n-1}{n-3} &= \frac{n}{S^2} Var(\bar{I}) \\
Var(\bar{I}) &= \frac{(n-1)S}{(n-2)n}
\end{align*}
[/tex]The distribution of the total number of registrations is the sum of the distributions of the registrations in the separate schools. According to the Central limit theorem, that sum is approximately normally distributed.
[tex]\bar{T} \sim N\left(n \cdot E(\bar{I}), n \cdot Var(\bar{I})\right)[/tex]
The standardized variable is:
[tex]\frac{\bar{T} - n \cdot E(\bar{I})}{\sqrt{n \cdot Var(\bar{I})}/ \sqrt{n}} \sim N(0,1) [/tex]
And hence the 95% confidence interval is:
[tex] \left[n \cdot E(\bar{I}) - z_{0.025} \sqrt{n \cdot Var(\bar{I})}/ \sqrt{n}, n \cdot E(\bar{I}) + z_{0.025} \sqrt{n \cdot Var(\bar{I})}/ \sqrt{n} \right] [/tex]
Plugging in [itex]n=2700, E(\bar{I})=3700, Var(\bar{I})=161441.4414[/itex] finally gives:
[tex] \left[ 9989212.476 , 9990787.524 \right] [/tex]
The solution manual gives an interval that is much wider than this.