Question

A county clerk wants to estimate the proportion of voters who will need special election facilities.
Suppose a sample of 400 voters was taken.
If 150 need special election facilities, what is the upper confidence limit (UCL) for the 90% confidence interval for the population proportion of voters who will need special election facilities?
Round your answer to 3 decimal places.

Answer 1

Answer: The upper confidence limit for the 90% confidence interval would be 0.415.

Step-by-step explanation:

Since we have given that

n = 400

x = 150

So, $\hat{p}=\dfrac{x}{n}=\dfrac{150}{400}=0.375$

At 90% confidence interval, z = 1.645

So, margin of error would be

$z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.375\times 0.625}{400}}\\\\=0.0398$

So, the upper limit would be

$\hat{p}+0.0398\\\\=0.375+0.0398\\\\=0.415$

Hence, the upper confidence limit for the 90% confidence interval would be 0.415.