Question

using a random sample of 1432 tv households, acme media statistics found that 61.1% watched the final episode of still hungry 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 hungry 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 for a proportion

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 \(p=0.617\) (since \(61.7\%=0.617\)) and \(n = 1432\).

Step2: Calculate \(1-p\)

\(1 - p=1 - 0.617 = 0.383\)

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

\(\frac{p(1 - p)}{n}=\frac{0.617\times0.383}{1432}=\frac{0.236311}{1432}\approx0.000165\)

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

\(\sqrt{0.000165}\approx0.0128\)

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

\(E=z\sqrt{\frac{p(1 - p)}{n}}=1.96\times0.0128 = 0.025088\approx 2.51\%\)

Step6: Write the confidence - interval statement

The confidence interval for the population proportion \(p\) is \(\hat{p}-E

Answer:

a. The margin of error is \(2.51\%\)
b. We are \(95\%\) confident that the percentage of TV households in the population who tuned into the final episode of "Still Hungry On" is between \(59.2\%\) and \(64.2\%\)