In each of Exercises $11.25-11.30$, we have given the number of successes and the sample size for a simple random sample from a population. In each case, do the following tasks.
a. Determine the sample proportion.
b. Decide whether using the one-proportion $z$-interval procedure is appropriate.
c. If appropriate, use the one-proportion $z$-interval procedure to find the confidence interval at the specified confidence level.
d. If appropriate, find the margin of error for the estimate of $p$ and express the confidence interval in terms of the sample proportion and the margin of error:
11.30 $x=3,n=100,99\%$level

To determine the sample proportion for $x=3,n=100,99\%$ level.

Let, the sample size $n$is $100$.

Then, the number of success $x$is $3$.

Determine the Sample proportion by:
$\hat{p}=\frac{x}{n}$
$=\frac{3}{100}$
$=0.03$
As a result, the sample proportion is $0.03$.

To decide whether using the one-proportion $z$-interval procedure is appropriate or not.

The following are the assumptions for one proportion $z$-interval procedure:

$np\ge 10$and $n(1-p)\ge 10$

$=3<10$

Also,

$n(1-p)=100(1-0.03)$

$=97>10$

One of the conditions is not satisfied in this case.

As a result, applying the one proportion $z$- interval is not appropriate.

To appropriate, use the one-proportion $z$-interval procedure to find the confidence interval at the specified confidence level.

It is clear from part (b) that the one-proportion $z$-interval procedure is ineffective.
As a result, the confidence interval cannot be calculated.

To appropriate, find the margin of error for the estimate of $p$and express the confidence interval in terms of the sample proportion and the margin of error:

It is clear from part (b) that the one-proportion $z$ interval procedure is ineffective.
As a result, the margin of error cannot be calculated.