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We are interested in estimating the proportion of students at a university who smoke. Out of a random sample of 200 students from this university, 40 students smoke. (a) Calculate a \(95 \%\) confidence interval for the proportion of students at this university who smoke, and interpret this interval in context. (Reminder: Check conditions.) (b) If we wanted the margin of error to be no larger than \(2 \%\) at a \(95 \%\) confidence level for the proportion of students who smoke, how big of a sample would we need?

Short Answer

Expert verified
(a) CI is approx. (0.1446, 0.2554); (b) need at least 1539 students.

Step by step solution

01

Check Conditions for Confidence Interval

Before calculating the confidence interval, we need to ensure that certain conditions are met for a normal approximation. The sample must be random, and both \(np\) and \(n(1-p)\) must be greater than 5.Here, \(n = 200\) and the sample proportion \(\hat{p} = \frac{40}{200} = 0.2\).Checking conditions: \(np \geq 5\) \(\Rightarrow 200 \times 0.2 = 40\), and \(n(1-p) \geq 5\) \(\Rightarrow 200 \times 0.8 = 160\).Both conditions are satisfied since 40 and 160 are both greater than 5.
02

Calculate the Confidence Interval

To calculate the confidence interval, use the formula:\[\hat{p} \pm Z \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} }\]Here, \(\hat{p} = 0.2\), \(n = 200\), and \(Z\) for a \(95\%\) confidence level is \(1.96\).Calculate the standard error (SE):\[ SE = \sqrt{ \frac{0.2 \times 0.8}{200} } = \sqrt{ \frac{0.16}{200} } \approx 0.02828 \]Then the confidence interval is:\[0.2 \pm 1.96 \times 0.02828 \approx (0.1446, 0.2554)\]
03

Interpret the Confidence Interval

The \(95\%\) confidence interval for the proportion of students who smoke is approximately \((0.1446, 0.2554)\). This means we are \(95\%\) confident that the true proportion of students who smoke at this university is between \(14.46\%\) and \(25.54\%\).
04

Determine Sample Size for Desired Margin of Error

We want the margin of error (ME) to be no larger than \(2\% = 0.02\) at a \(95\%\) confidence level. The margin of error is given by the formula:\[ ME = Z \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \]Setting \(ME = 0.02\) and \(Z = 1.96\), we have:\[ 0.02 = 1.96 \sqrt{ \frac{0.2 \times 0.8}{n} } \]Rearrange to solve for \(n\):\[ \sqrt{ \frac{0.16}{n} } = \frac{0.02}{1.96} \approx 0.0102 \]\[ \frac{0.16}{n} = 0.0102^2 \approx 0.000104 \]\[ n = \frac{0.16}{0.000104} \approx 1538.46 \]Thus, \(n\) must be at least 1539 when rounded to the nearest whole number.
05

Conclusion for Sample Size

To achieve a margin of error no larger than \(2\%\) at a \(95\%\) confidence level, you would need a sample size of at least 1539 students.

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Key Concepts

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

Sample Size Determination
Determining the appropriate sample size is crucial when designing surveys or studies, especially when you want your results to be reliable and within a certain level of accuracy. When talking about determining sample size for a confidence interval, you need to know the desired margin of error and confidence level. These parameters help you calculate how many observations you need to make sure your sample accurately reflects the population.

To find out the sample size for a given margin of error in a proportion estimate, you use the formula:
  • The margin of error (\( ME \)) is given by \( ME = Z \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \), where \( \hat{p} \) is the sample proportion, \( n \) is the sample size, and \( Z \) is the Z-score for your confidence level.
  • Rearrange the formula to solve for the sample size \( n \): \[ n = \frac{Z^2 \hat{p}(1-\hat{p})}{ME^2} \]
This allows researchers to decide how large their sample must be to achieve statistical significance. An inadequate sample size can lead to unreliable results, while too large a sample might be inefficient.
Margin of Error
The margin of error is a measure of how much you could expect your sample's proportion to differ from the true population proportion. It gives a range about the sample proportion that likely includes the true proportion, explaining the potential error in your estimate.

When calculating confidence intervals, the margin of error is determined using the formula:
  • The margin of error (\( ME \)) is expressed by \( ME = Z \times SE \), where \( SE \) is the standard error: \( \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \).
  • The Z-score (\( Z \)) corresponds to the chosen confidence level, which for a 95% confidence interval is typically 1.96.
The margin of error decreases as the sample size increases. If the sample size is larger, the sample proportion is a more precise estimate of the population proportion. This is why researchers often aim to achieve a desired margin of error when determining sample size.
Normal Approximation
The normal approximation is a statistical method allowing us to calculate confidence intervals and perform hypothesis testing on sample data. It's necessary because in many cases, directly calculating the exact probability can be complex. Instead, we approximate the sample distribution using a normal distribution, simplifying calculations.

The conditions for using a normal approximation are:
  • The sample size, \( n \), must be large enough. This rule of thumb is that both \( np \) and \( n(1-p) \) should be greater than 5.
  • This ensures the distribution is approximately normal, making the approximation valid.
When these conditions are met, you can use the normal approximation to construct a confidence interval for the sample proportion, providing a straightforward way to infer population parameters from sample data.
Sample Proportion
The sample proportion (\( \hat{p} \)) is a statistic that estimates the population proportion (\( p \)). It is calculated by dividing the number of successful outcomes (such as smokers) by the total number of observations in the sample.

In our exercise, the sample proportion was calculated as follows:
  • There were 40 students who smoke out of 200, thus \( \hat{p} = \frac{40}{200} = 0.2 \).
This proportion is then used in other calculations, such as determining confidence intervals or calculating the sample size for a desired margin of error. It forms the basis of many statistical analyses and is a crucial element when interpreting study results.

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Most popular questions from this chapter

A physical education teacher at a high school wanting to increase awareness on issues of nutrition and health asked her students at the beginning of the semester whether they believed the expression "an apple a day keeps the doctor away", and \(40 \%\) of the students responded yes. Throughout the semester she started each class with a brief discussion of a study highlighting positive effects of eating more fruits and vegetables. She conducted the same apple-a- day survey at the end of the semester, and this time \(60 \%\) of the students responded yes. Can she used a two-proportion method from this section for this analysis? Explain your reasoning.

A national survey conducted among a simple random sample of 1,507 adults shows that \(56 \%\) of Americans think the Civil War is still relevant to American politics and political life. \({ }^{47}\) (a) Conduct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant. (b) Interpret the p-value in this context. (c) Calculate a \(90 \%\) confidence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the interval in this context, and comment on whether or not the confidence interval agrees with the conclusion of the hypothesis test.

Researchers conducted a study investigating the relationship between caffeinated coffee consumption and risk of depression in women. They collected data on 50,739 women free of depression symptoms at the start of the study in the year 1996 , and these women were followed through 2006 . The researchers used questionnaires to collect data on caffeinated coffee consumption, asked each individual about physician- diagnosed depression, and also asked about the use of antidepressants. The table below shows the distribution of incidences of depression by amount of caffeinated coffee consumption. \(^{52}\) (a) What type of test is appropriate for evaluating if there is an association between coffee intake and depression? (b) Write the hypotheses for the test you identified in part (a). (c) Calculate the overall proportion of women who do and do not suffer from depression. (d) Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (Observed - Expected) \(^{2} /\) Expected. (e) The test statistic is \(\chi^{2}=20.93\). What is the p-value? (f) What is the conclusion of the hypothesis test? (g) One of the authors of this study was quoted on the NYTimes as saying it was "too early to recommend that women load up on extra coffee" based on just this study. \(^{53}\) Do you agree with this statement? Explain your reasoning.

The General Social Survey asked 1,578 US residents: "Do you think the use of marijuana should be made legal, or not?" \(61 \%\) of the respondents said it should be made legal. 13 (a) Is \(61 \%\) a sample statistic or a population parameter? Explain. (b) Construct a \(95 \%\) confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data. (c) A critic points out that this \(95 \%\) confidence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain. (d) A news piece on this survey's findings states, "Majority of Americans think marijuana should be legalized." Based on your confidence interval, is this news piece's statement justified?

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