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The normal distribution is a crucial concept in statistics, particularly in finding confidence intervals. Imagine the shape of a symmetric bell curve, where most data points cluster around the mean. This is the normal distribution, and it's used to estimate probabilities.
It helps create a confidence interval, which is a range of values used to estimate a population parameter with a specific level of confidence. In this problem, we want to find a confidence interval for the difference between two proportions, namely \(p_1\) and \(p_2\).

• The normal distribution allows the use of z-scores, which tell us how many standard deviations a data point is from the mean.
• A 99% confidence interval corresponds to a z-score of approximately 2.58. This z-score indicates that we are 99% confident that the true difference lies within the calculated interval.

By using the properties of the normal distribution, we can determine how likely it is for sample data to reflect the population, enabling a better estimation of the true difference in proportions.

The difference in proportions is a fundamental step in this exercise. It represents the difference between the sample proportion of Group 1 \(p_1 = 0.76\) and that of Group 2 \(p_2 = 0.9\).
To find the difference:

• Subtract one proportion from the other: \(p_1 - p_2\).
• The best estimate for this difference is simply the result from this subtraction.

In the given problem, this calculation yields \(-0.14\). This means Group 2 has a higher proportion of yes responses by 14% compared to Group 1. Understanding this difference is crucial, as it forms the basis for further calculations like the standard error and margin of error, ultimately affecting the confidence interval outcome.

The margin of error is vital in crafting a confidence interval. It helps gauge the uncertainty associated with the sample estimate.
To calculate the margin of error in this context:

• First, determine the standard error (SE) of the difference in proportions. The SE considers the variability in each sample and combines it.
• Use the formula: \(SE = \sqrt{(p_1(1 - p_1) / n_1) + (p_2(1 - p_2) / n_2)}\).
• Then, multiply the SE by the z-score associated with the desired confidence level. For a 99% confidence interval, the z-score is 2.58.

In this exercise, the standard error was computed to be \(0.053\), and the margin of error was found to be \(0.137\). This margin of error means that while our best estimate of the difference in proportions is \(-0.14\), it could reasonably vary by as much as \(0.137\) on either side.
Therefore, the confidence interval is \([-0.277, -0.003]\), indicating where the true difference in population proportions likely falls. The margin of error acknowledges the sample's inherent randomness, providing a buffer zone around the point estimate.