Student body totals Use your interval from Exercise 40 to construct and interpret a $99\%$confidence interval for the total number of students at the school that the student body president can identify by name. Then use your interval to evaluate the president’s claim.

It is given that

$c=99\%=0.99$

$x=46$

$n=100$

$1-\alpha =0.99$, using probability table$z_{a/2}=2.575$

The conditions are:

Random: Sample are chosen randomly, so this condition is satisfied.

Independent: Sample of $100$student chips is surely less than $10\%$of all population of role="math" localid="1654452833197" $1800$students.

Normal: Number of success is $46$and failures are $100-46=54$. This is greater than $10$.

All conditions are satisfied.

Sample Proportion: $\hat{p}=\frac{x}{n}$

$=\frac{46}{100}=0.46$

Margin of Error: $E=z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=2.575\times \sqrt{\frac{0.46(1-0.46)}{100}}=0.1283$

Confidence Interval: $\hat{p}-E=0.46-0.1283=0.3317$

$\hat{p}+E=0.46+0.1283=0.5883$

As total students are $1800$

$1800(0.3317)=597.06\approx 597$

$1800(0.5883)=1058.94\approx 1059$

Hence, number of students that student body present knows lie between$597-1059$