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TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device. Calculate a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time. What is the upper bound of this 99% confidence level for the proportion

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Answer:

99% Confidence interval:  (0.5034,0.5566)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 2343

Proportion of adults who watched digitally streamed TV = 53\%

[tex]\hat{p} = 0.53[/tex]

99% Confidence interval:

[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58[/tex]

Putting the values, we get:

[tex]0.53\pm 2.58(\sqrt{\dfrac{0.53(1-0.53)}{2343}})\\\\ = 0.53\pm 0.0266\\\\=(0.5034,0.5566)\\\approx (0.50, 0.56)[/tex]

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