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Chapter 7: Problem 20

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. a. Calculate and interpret 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. b. What sample size would be required for the width of a \(99 \%\) CI to be at most. 05 irrespective of the value of \(\hat{p}\) ?

Short Answer

Expert verified
a. The 99% CI is (50.35%, 55.65%). b. A sample size of 2658 is needed.

Step by step solution

01

Identify Parameters

First, understand what needs to be calculated. We need a 99% confidence interval for the proportion of American adults who have watched digitally streamed TV programming. Key information:- Sample size \( n = 2343 \)- Sample proportion \( \hat{p} = 0.53 \)- Confidence level = 99% (implies \( z \approx 2.576 \) for 99% confidence).
02

Calculate Standard Error

The standard error of the proportion is calculated using the formula \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Substitute the known values:\[ SE = \sqrt{\frac{0.53(1-0.53)}{2343}} \].Calculate to find \( SE \approx 0.0103 \).
03

Determine Margin of Error

The margin of error (ME) is calculated using the formula \( ME = z \cdot SE \). With \( z = 2.576 \) for 99% confidence:\[ ME = 2.576 \times 0.0103 \].Calculate to find \( ME \approx 0.0265 \).
04

Construct Confidence Interval

The confidence interval is given by \( \hat{p} \pm ME \). Substitute the known values:\[ CI = 0.53 \pm 0.0265 \].This results in the interval (0.5035, 0.5565).
05

Interpret Confidence Interval

We are 99% confident that the proportion of all adult Americans who have watched digitally streamed TV programming falls between 50.35% and 55.65%.
06

Determine Required Sample Size for Given Width

To find the required sample size for a CI width of 0.05 at 99% confidence, use the formula:\[ n = \left( \frac{z^2 \times p \times (1-p)}{E^2} \right) \], where \( E \) is the desired margin of error (0.025 for half-width 0.05).Assume \( p = 0.5 \) for maximum variance:\[ n = \left( \frac{2.576^2 \times 0.5 \times 0.5}{0.025^2} \right) \].Calculate to find \( n \approx 2658 \).
07

Conclusion

For part (b), a sample size of at least 2658 is required for the 99% confidence interval to have a maximum width of 0.05, regardless of the estimated proportion \( \hat{p} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Size Calculation
Determining the right sample size is crucial in statistics to ensure that our results are both accurate and reliable. In this exercise, the goal was to find the sample size needed for a confidence interval to have a specific width at 99% confidence. The required formula for this is \[ n = \left( \frac{z^2 \times p \times (1-p)}{E^2} \right) \]. This formula helps us understand that our sample size (\( n \)) depends on the:
  • z-score, associated with our desired confidence level (for 99%, it is approximately 2.576)
  • Estimated proportion (\( p \)), which in situations with unknown proportions is best to use 0.5 for maximum variability
  • Desired margin of error (\( E \)), which is half the width of the confidence interval
Using these parameters ensures our study is both practical and cost-effective, providing the best estimate with the resources at hand.
Margin of Error
The margin of error (ME) is a crucial concept in statistical analysis, representing the range of values above and below the sample proportion within which we expect the real population proportion to lie. It helps us express the level of uncertainty in our estimates, giving a boundary around our statistic. ME is calculated with the formula \( ME = z \cdot SE \), where \( z \) is your z-score depending on the confidence level, and \( SE \) represents the standard error. For instance, in our exercise, using a confidence level of 99%, the z-score is 2.576. Multiplying this by the \( SE \) provides the \( ME \) of approximately 0.0265. Understanding the ME is vital for interpreting results properly and informs us about the range within which the true population parameter likely resides.
Standard Error
The standard error (SE) quantifies the variability of the sample statistic estimated from different samples. It essentially measures how much the sample proportion is expected to fluctuate from the true population proportion due to random sampling errors in different samples. The formula used for calculating SE of a proportion is \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion and \( n \) is the sample size. In this scenario, with a sample size of 2343 and a sample proportion of 0.53, we found \( SE \approx 0.0103 \). The smaller the standard error, the closer your sample proportion is to the true population proportion, implying a more precise estimate.
Proportion Estimation
Proportion estimation is a statistical technique used to infer about a particular characteristic within a whole population, based on a sample proportion. In our worked exercise, the sample proportion \( \hat{p} \) was determined by dividing the number of survey participants who watched streamed content by the total number surveyed, resulting in \( \hat{p} = 0.53 \) or 53%. To infer about the true proportion of adult Americans watching streamed TV programming, we utilize properties of proportions alongside standard errors, margin of error, and confidence intervals. Recognizing and estimating the correct proportion is critical for making informed conclusions about population behaviors and trends, which assists in decision making across various fields.

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