Which would be hardest for you to give up: Your computer or your television? In a survey of 1677 U.S. Internet users, \(74 \%\) of the young tech elite (average age of 22 ) say their computer would be very hard to give up (PC Magazine, February 3, 2004). Only 48\% say their television would be very hard to give up. a. Develop a \(95 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer. b. Develop a \(99 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their television. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.

First, calculate the sample proportion (p̂) for the young tech elite that would find it very hard to give up their computer. The sample size is 1677, and 74% say their computer would be very hard to give up. The sample proportion can be calculated as follows: \(p̂ = \frac{x}{n}\) where \(x\) is the number of young tech elite who would find it very hard to give up their computer, and \(n\) is the total number of U.S. Internet users surveyed. \(p̂ = \frac{0.74 \times 1677}{1677} = 0.74\)

Next, we will calculate the margin of error (ME) for the 95% confidence interval. To do this, we first need to find the standard deviation, which was calculated using the following formula: \(standard deviation = \sqrt{\frac{p̂(1-p̂)}{n}}\) Now, we can calculate the margin of error: \(ME = z \times standard deviation\) For a 95% confidence interval, the z-value (critical value) is 1.96. We plug in the values and get the margin of error as: \(ME = 1.96 \times \sqrt{\frac{0.74(1-0.74)}{1677}} = 0.0196\) Finally, we find the confidence interval: \(CI = (p̂ - ME, p̂ + ME) = (0.74 - 0.0196, 0.74 + 0.0196) = (0.7204, 0.7596)\) The 95% confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer is (0.7204, 0.7596).

We follow the same steps as part (a) and begin by calculating the sample proportion (p̂) for the young tech elite that would find it very hard to give up their television. The sample size is 1677, and 48% say their television would be very hard to give up: \(p̂ = \frac{0.48 \times 1677}{1677} = 0.48\)

To calculate the margin of error (ME) for the 99% confidence interval, we will need to find the standard deviation using the same formula as in part (a): \(standard deviation = \sqrt{\frac{p̂(1-p̂)}{n}}\) Now, we calculate the margin of error: \(ME = z \times standard deviation\) For a 99% confidence interval, the z-value (critical value) is 2.576. We plug in the values and get the margin of error as: \(ME = 2.576 \times \sqrt{\frac{0.48(1-0.48)}{1677}} = 0.0260\) Finally, we find the confidence interval: \(CI = (p̂ - ME, p̂ + ME) = (0.48 - 0.0260, 0.48 + 0.0260) = (0.454, 0.506)\) The 99% confidence interval for the proportion of the young tech elite that would find it very hard to give up their television is (0.454, 0.506).

Now we will compare the margin of errors of part (a) and part (b): \(ME_{Computers} = 0.0196\) \(ME_{Television} = 0.0260\) The margin of error is larger for part (b) (Television). The reason behind this is twofold: 1. The confidence level is higher in part (b) (99%) compared to part (a) (95%). As the confidence level increases, the margin of error also increases, in order to provide a larger range for our estimation. 2. The proportions are different in part (a) and part (b) (0.74 vs. 0.48). This means that the standard deviation for the two groups would be different, as it is dependent on the proportion. This difference in standard deviation could also contribute to the difference in the margin of error.