Question

using a random sample of 1412 tv households, acme media statistics found that 51.1% 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 turned 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: Recall the formula for margin of error

The formula for margin of error \(E\) for a proportion is \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a \(95\%\) confidence level, \(z = 1.96\). Here \(p=0.511\) and \(n = 1412\).

Step2: Calculate \(1-p\)

\(1 - p=1 - 0.511=0.489\)

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

\(\frac{0.511\times0.489}{1412}=\frac{0.25}{1412}\approx0.000177\)

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

\(\sqrt{0.000177}\approx0.0133\)

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

\(E=1.96\times0.0133 = 0.0261\) or \(2.61\%\)

Answer:

a. The margin of error is \(2.61\%\)

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 \((51.1 - 2.61)\%\) and \((51.1+ 2.61)\%\), that is between \(48.49\%\) and \(53.71\%\)