Question

using a random sample of 1412 tv households, acme media statistics found that 61.1% watched the final episode of still hanging on.
a. find the margin of error in this percent.
b. write a statement about the percentage of tv households in the population who tuned into the final episode of still hanging on.
a. the margin of error is ±□
(do not round until the final answer. then round to the nearest hundredth as needed.)

Explanation:

Step1: Identify the formula for margin of error

For a proportion, the margin of error \(E\) is given by \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a 95% confidence level (common in such problems when not specified otherwise), \(z = 1.96\). Here \(p=0.611\) and \(n = 1412\).

Step2: Calculate \(1-p\)

\(1 - p=1 - 0.611=0.389\)

Step3: Calculate \(\frac{p(1 - p)}{n}\)

\(\frac{0.611\times0.389}{1412}=\frac{0.237679}{1412}\approx0.0001683\)

Step4: Calculate \(\sqrt{\frac{p(1 - p)}{n}}\)

\(\sqrt{0.0001683}\approx0.01297\)

Step5: Calculate the margin of error \(E\)

\(E = 1.96\times0.01297\approx0.0254\)

Answer:

a. The margin of error is \(\pm2.54\%\)
b. We are 95% confident that the percentage of TV households in the population who tuned into the final episode of "Still Hanging On" is between \(61.1\% - 2.54\%=58.56\%\) and \(61.1\%+ 2.54\% = 63.64\%\)