Question

In 1991, California imposed a "snack tax" (a sales \(\operatorname{tax}\) on snack food) in an attempt to help balance the state budget. A proposed alternative tax was a \(12 \phi\) -per-pack increase in the cigarette tax. In a poll of 602 randomly selected California registered voters, 445 responded that they would have preferred the cigarette tax increase to the snack tax (Reno Gazette-Journal, August 26,1991 ). Estimate the true proportion of California registered voters who preferred the cigarette tax increase; use a \(95 \%\) confidence interval.

Short answer

The 95% confidence interval for the true proportion of California registered voters who preferred the cigarette tax increase is from 0.704 to 0.774.

Step-by-step solution

Calculate the sample proportion

The sample proportion (p) is calculated by dividing the number of respondents that preferred the cigarette tax increase by the total number of respondents. So, \( p = \frac{445}{602} = 0.739 \).

Determine the z score for the given confidence level

The given confidence level is 95%. This corresponds to a z score of 1.96 (you can find this value from standard normal distribution tables). So, \( z = 1.96 \).

Calculate the standard error

The standard error (SE) is calculated using the formula: \( SE = \sqrt{ \frac{(p \cdot (1-p))}{n} } \), where n is the number of respondents. So, \( SE = \sqrt{ \frac{(0.739 \cdot (1-0.739))}{602} } = 0.018 \).

Calculate the confidence interval

The confidence interval is calculated using the formula: \( CI = p \pm (z \cdot SE) \). So, the confidence interval is \( CI = 0.739 \pm (1.96 \cdot 0.018) = 0.739 \pm 0.035 \), which means the interval is from 0.704 to 0.774.

Key concepts

Sample Proportion

Understanding the concept of a sample proportion is crucial in statistics, especially when we need to make inferences about a population. A sample proportion represents the fraction of individuals in a sample that exhibit a certain characteristic or preference. For instance, in the provided exercise, residents of California were asked if they preferred a cigarette tax increase over a snack tax. Out of 602 individuals surveyed, 445 preferred the cigarette tax increase. Therefore, the sample proportion (\( p \)) is calculated by dividing the number of affirmative responses by the total number of responses: \( p = \frac{445}{602} = 0.739 \).

This proportion indicates that approximately 73.9% of the sampled voters prefer the cigarette tax increase. This statistic is a snippet of information, but it becomes powerful when we use it as a basis to estimate the preferences of all California voters—a much larger group than our sample.

Standard Error

When we state a sample proportion, it is important to communicate how precise that estimate is. This is where standard error (SE) comes in. It is a measure of the statistical accuracy of an estimate. To calculate standard error, we use the formula: \( SE = \sqrt{ \frac{(p \/cdot (1-p))}{n} } \), where \( p \) is the sample proportion and \( n \) is the sample size.

In the exercise, the standard error helps us understand the variability or uncertainty in the proportion of voters who prefer the cigarette tax increase. The smaller the standard error, the more confident we can be that our sample proportion is close to the true population proportion. The calculated standard error in this case is \(\textbf{0.018}\), which allows us to gauge the potential error in estimating the true preference of all California voters based on our sample.

Normal Distribution

The normal distribution is a cornerstone of statistical analysis, often referred to as the bell curve for its characteristic shape. It's a continuous probability distribution symmetrical around the mean, where data tends to cluster, and less common extremes taper off on both ends.

In the context of confidence intervals, we assume that the sampling distribution of the sample proportion is approximately normal, especially as the sample size grows larger, thanks to the Central Limit Theorem. The z-score mentioned in the example (\( z = 1.96 \) for a 95% confidence level) is a value from the standard normal distribution that indicates the number of standard deviations a certain value is away from the mean. By using the z-score and the standard error, we effectively create a range (confidence interval) that is likely to capture the true population proportion, assuming that the distribution of sample proportions approximates a normal distribution.

Specifically, for our California voter example, the z-score tells us how far from the sample proportion we must go to be 95% sure we've included the true population proportion. The resulting confidence interval extends from \( 0.704 \) to \( 0.774 \) which gives us a range of values that likely includes the actual preference level of all voters regarding the cigarette tax increase.