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Chapter 3: Problem 118

How Important Is Regular Exercise? In a recent poll \(^{50}\) of 1000 American adults, the number saying that exercise is an important part of daily life was 753 , Use StatKey or other technology to find and interpret a \(90 \%\) confidence interval for the proportion of American adults who think exercise is an important part of daily life.

Short Answer

Expert verified
The 90% confidence interval for the proportion of American adults who think exercise is an important part of daily life is approximately (0.730, 0.776). This suggests with 90% confidence that the true proportion in the total population falls between these two values. This means that 73% to 77.6% of American adults might think that exercise is important for daily life.

Step by step solution

01

Calculate Sample Proportion

The sample proportion (p) is obtained by dividing the number of people who agreed with the statement by the total number of people questioned, that is \(p = \frac{753}{1000} = 0.753\).
02

Find the Standard Deviation of Proportion

The standard deviation \(\sigma\) of the sample proportion can be found using the formula \(\sigma = \sqrt{(p \cdot (1-p))/n}\), so \(\sigma = \sqrt{(0.753 \cdot (1-0.753))/1000} = 0.01378\).
03

Determine Z-Score for 90% Confidence Interval

For a 90% confidence interval, the z-score (which is the number of standard deviations a data point is from the mean) can be taken from statistical tables or using a statistical tool to be approximately 1.645.
04

Calculate Confidence Interval

The 90% confidence interval can be calculated as (\(p - Z \cdot \sigma\), \(p + Z \cdot \sigma\))= (0.753 - 1.645 \cdot 0.01378, 0.753 + 1.645 \cdot 0.01378) = (0.730, 0.776).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation of Proportion
When dealing with percentages or proportions in statistics, understanding the concept of standard deviation of a proportion is essential. It's a measure of the variability of a sample proportion, indicating how much the proportion is expected to fluctuate due to random sampling error.

The formula for the standard deviation of a proportion is \( \sigma = \sqrt{\frac{p(1-p)}{n}} \) where \( p \) is the sample proportion and \( n \) is the sample size. In our poll example, we found that 75.3% of the 1000 adults surveyed think exercise is an important part of daily life. Using this sample proportion, we calculated the standard deviation to be \( \sigma = 0.01378 \). This small standard deviation indicates that there is relatively little variability around the sample proportion.
Sample Proportion
The sample proportion is a statistic that provides an estimate of the true proportion in the entire population. It's calculated simply by dividing the number of individuals in the sample with a certain characteristic by the total number of individuals in the sample. In our exercise scenario, \( p = \frac{753}{1000} = 0.753 \), meaning that approximately 75.3% of the sampled American adults believe that exercise is important.

Understanding the sample proportion is fundamental because it serves as the starting point for constructing confidence intervals. These intervals offer a range in which we believe the true population proportion lies, with a certain level of confidence.
Z-Score for Confidence Interval
The z-score, in the context of a confidence interval, helps us determine how many standard deviations we should go from the sample proportion to capture the center of the population proportion with a specific confidence level. For our 90% confidence interval, the z-score is approximately 1.645. This value means that we are looking at 1.645 standard deviations from the sample proportion on each side to estimate the range for the true population proportion.

To obtain our confidence interval, we multiply our standard deviation \( \sigma \) by the z-score and then both add and subtract this product from the sample proportion \( p \). Concretely, for the exercise, the range \((0.730, 0.776)\) indicates with 90% confidence that the proportion of all American adults who consider exercise essential falls between 73% and 77.6%. This interval does not assert that precisely 90% of adults fall within this range but rather, if we were to take many samples and calculate intervals in the same way, about 90% of those intervals would contain the true population proportion.

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Most popular questions from this chapter

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