91Ó°ÊÓ

To determine the sample proportion for $x=16,n=20,90\%$ level.

Let, the sample size $n$is $20$
And the number of successes $x$is $16$.
Determine the sample proportion by:
$\hat{p}=\frac{x}{n}$
$=\frac{16}{20}$
$=0.8$
Asa a result, the sample proportion is $0.8$.

To decide whether using the one-proportion $z$-interval procedure is appropriate.

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

A simple random sampling method should be used to select the sample.
The number of successes $x$and failures$n-x$ should both be at least five.
The number of successes in this case, $x=16$, is larger than $5$.
The following formula is used to calculate the number of failures:
$n-x=20-16$
$=4$

The number of failures is less than $5$in this case.

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.