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A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that \(55 \%\) of high school seniors are fairly certain that they will participate in a study abroad program in college. \(^{19}\) (a) Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning. (b) Let's suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a \(90 \%\) confidence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context. (c) What does "90\% confidence" mean? (d) Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college?

Short Answer

Expert verified
(a) No, it's likely biased. (b) 90% CI is (0.5309, 0.5691). (c) 90% of similar samples contain the true proportion. (d) Yes, majority is supported.

Step by step solution

01

Evaluate the sample representativeness

To determine if this sample is representative, consider whether the way participants were selected provides a fair representation of all high school seniors in the US. Since only seniors who took the SAT and completed the optional web survey are included, the sample is likely biased. Not all high school seniors take the SAT, and those who completed the web survey might have different interests or characteristics than those who did not.
02

Calculate the standard error

The standard error for the proportion can be calculated using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p = 0.55\) is the sample proportion and \(n = 1509\) is the sample size. Substituting the values, we have \(SE = \sqrt{\frac{0.55 \times 0.45}{1509}}\).
03

Construct the confidence interval

For a 90% confidence interval, use the critical value \(z^* = 1.645\). The confidence interval is given by \(p \pm z^* \times SE\). Substituting the values, we have \(0.55 \pm 1.645 \times SE\). This calculates to approximately \(0.5309\) and \(0.5691\). So, the 90% confidence interval is \((0.5309, 0.5691)\).
04

Interpret the confidence interval

The 90% confidence interval \((0.5309, 0.5691)\) suggests that we are 90% confident that the true proportion of high school seniors fairly certain about studying abroad is between 53.09% and 56.91%. This range allows us to estimate the population proportion based on our sample.
05

Explain 90% confidence

A 90% confidence level means that if we were to take many random samples and construct an interval from each one, about 90% of those intervals would contain the true population proportion. It reflects the reliability of our sampling method.
06

Assess the claim of majority

The confidence interval \((0.5309, 0.5691)\) does not contain 50%, meaning we can be 90% confident that more than half of the seniors are certain they want to study abroad, supporting the claim that a majority are fairly certain.

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Key Concepts

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

Sample Representativeness
When assessing "sample representativeness," it is crucial to consider whether the sampled individuals accurately reflect the entire population. In this exercise, the sample consists only of high school seniors who took the SAT and filled out an optional survey. This significant restriction indicates that the sample might not represent all high school seniors across the U.S.
Not all seniors take the SAT, and those who volunteer for web surveys may have unique characteristics or interests that do not align with those who did not participate in the survey.
This leads to concerns about bias, where the opinions and characteristics of the sample do not accurately reflect those of the broader population. To mitigate such issues, random sampling from a more inclusive pool would enhance representativeness.
Population Proportion
The "population proportion" refers to the fraction of the total population that exhibits a particular trait or behavior. In the context of the exercise, we are interested in the proportion of all high school seniors who are fairly certain they will participate in a study abroad program in college.
The sample proportion ( p ext{=0.55} ) is used as an estimate of the population proportion. Although it provides valuable insight, it is important to recognize that some degree of uncertainty is inherently involved, primarily due to the sample's non-representative nature.
Through constructing a confidence interval, we aim to provide a range of plausible values for the actual population proportion, allowing for a more informed inference about the broader population.
Standard Error
The "standard error" is a measure of the variability or precision of the sample proportion in estimating the true population proportion. It provides an indication of how much the sample proportion (p ext{=0.55}) might differ from the actual population proportion.
It is calculated using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] where p ext{=0.55} represents the sample proportion and ext{n=1509} the sample size.
In this case, the calculated standard error provides a basis for constructing the confidence interval, showing the expected range in which the true population proportion is likely to fall. The smaller the standard error, the more precise our confidence interval becomes.
Critical Value
The "critical value" is a key determinant when calculating a confidence interval, representing the number of standard errors to move away from the sample proportion to achieve a desired confidence level.
In this exercise, for a 90% confidence level, the critical value is ext{z* = 1.645} . This value is derived from the standard normal distribution and signifies the cut-off point beyond which lies 10% of the distribution (5% on each end).
The critical value ensures that approximately 90% of confidence intervals, constructed from various samples of the same size, would contain the true population proportion. Therefore, it guarantees a relatively high degree of reliability in the estimation process shown by the resulting confidence interval.

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Most popular questions from this chapter

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