91Ó°ÊÓ

The sample proportion is a crucial part of statistics, especially when estimating the true proportion of a population. It is represented as \(\hat{p}\). In this exercise, the sample proportion was calculated by dividing the number of adults who always wear their seat belt (769) by the total number of adults surveyed (1010). This gives us: \[ \hat{p} = \frac{769}{1010} \approx 0.761 \] Essentially, \(\hat{p}\) tells us that approximately 76.1% of the sampled adults reported always wearing their seat belt.
Understanding the sample proportion is fundamental in estimating the behavior of the overall population based on our sample.

The confidence level tells us how sure we can be about our estimates. It is often expressed as a percentage. In this exercise, a 98% confidence level means that we are 98% sure the true proportion of adults who always wear their seat belt lies within our calculated interval. Higher confidence levels provide more assurance but result in wider intervals, while lower confidence levels give narrower intervals but with less certainty.
It's vital to balance confidence levels to keep the estimates both useful and believable.

The standard error (SE) measures the accuracy of our sample proportion as an estimate of the true population proportion. It accounts for sample size and variability within the data. The formula for standard error in a proportion context is: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \(\hat{p}\) is the sample proportion, and \( \ \) is the sample size.
For our exercise, substituting \(\hat{p} = 0.761 \) and \( n = 1010 \) into the formula, we calculated: \[ SE = \sqrt{\frac{0.761 \times 0.239}{1010}} \approx 0.014 \] The smaller the SE, the more accurate our sample proportion is as an estimate.

The z-score is a statistic that represents how many standard deviations our sample proportion is away from the mean population proportion under the assumption of the null hypothesis. For a given confidence level, we can find the corresponding z-score from standard normal distribution tables or using calculators.
In this exercise, the 98% confidence level has a corresponding z-score of approximately 2.33. This z-score was used to calculate the margin of error (ME) by multiplying it with the standard error: \[ ME = z \times SE \] With \( z = 2.33 \) and \( SE = 0.014 \), this gives: \[ ME = 2.33 \times 0.014 \approx 0.033 \] Finally, the margin of error helps construct our confidence interval by adding and subtracting it from \(\hat{p}\), resulting in our interval estimate.