Question

Nielsen ratings are based on televisions in 50005000 households. Nielsen estimates that 12 comma 00012,000 people live in these households. Suppose Nielsen reports that American Idol had 6565​% of the TV audience. Interpret this result with a​ 95% confidence​ interval, based on a sample size of 12 comma 00012,000 people.

Answer 1

Answer:

We are 95% confident that the true proportion of TV audience is between 65.15% and 65.85%

Step-by-step explanation:

-From the given information, $\hat p=0.65$.

-We calculate the confidence interval using this value at 95% confidence level:

$CI=\hat p\pm z \sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\\\=0.65\pm 1.96\times \sqrt{\frac{0.65\times 0.35}{12000}}\\\\\\=0.65\pm 0.0085\\\\\\=[0.6415,0.6585]$

So, the 95% confidence interval is (0.6515,0.6585).

Hence, we are 95% confident that the true proportion of TV audience is between 65.15% and 65.85%.