Final answer:
A 95% confidence interval for the proportion of all teens who have a TV set in their room is calculated using the sample proportion and the sample size, along with the z-score for the desired confidence level. The interval options provided suggest that the correct interval could be option c, which is centered around the sample proportion of approximately 0.66.
Step-by-step explanation:
To determine a 95% confidence interval for the proportion of all teens who have a TV set in their room, we use the sample proportion p and the sample size n to calculate the margin of error (ME) and then create the interval using the formula p ± ME. Given that 693 out of 1050 teenagers have a television in their room, the sample proportion p is 693/1050, which simplifies to approximately 0.66.
Using the standard formula for the margin of error at a 95% confidence level:
ME = z * sqrt((p * (1 - p)) / n)
Where z is the z-score corresponding to a 95% confidence level (1.96 for a two-tailed test), the calculation provides us with a numerical value for ME. We then apply this number to create the confidence interval around the sample proportion p. Without doing the calculation, the only option among the given choices that seems plausible, based on our computed sample proportion of ~0.66, is option c. This can be verified by actually computing the interval, which should yield a range on either side of 0.66 that is reasonable for a sample size of 1050.
Option a cannot be correct because the interval is out of the possible range for a proportion (0 to 1). Options b and d are close, but c seems to cover more of the potential variability while remaining centered around the computed sample proportion of 0.66.