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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) The 95% confidence interval is approximately (14.45%, 25.55%). (b) We need a sample size of 1537 for a 2% margin of error.

Step by step solution

01

Identify the Number of Smokers

From a sample of 200 students, 40 students smoke. We need to calculate the proportion of smokers in the sample.
02

Calculate Sample Proportion

Find the sample proportion of smokers by dividing the number of smokers by the sample size: \( \hat{p} = \frac{40}{200} = 0.2 \).
03

Check Conditions for Confidence Interval

Ensure that the sample size is large enough by checking that both \( n\hat{p} \) and \( n(1-\hat{p}) \) are greater than 5. Here, \( 200 \times 0.2 = 40 \) and \( 200 \times 0.8 = 160 \). Both values are greater than 5, so conditions are met.
04

Find the Z-value for 95% Confidence

For a 95% confidence interval, the Z-value is 1.96.
05

Calculate Standard Error

Compute the standard error using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.2 \times 0.8}{200}} \approx 0.0283 \).
06

Calculate Confidence Interval

The confidence interval is calculated using: \( \hat{p} \pm Z \times SE \). Therefore, the 95% confidence interval is \( 0.2 \pm 1.96 \times 0.0283 \approx (0.1445, 0.2555) \).
07

Interpret Confidence Interval

We are 95% confident that the true proportion of students who smoke at the university is between 14.45% and 25.55%.
08

Determine Sample Size for Smaller Margin of Error

If the margin of error needs to be 2% or 0.02, use the formula \( ME = Z \times SE \). Here, \( 0.02 = 1.96 \times \sqrt{\frac{p(1-p)}{n}} \). Solving for \( n \), we use an estimated \( p = 0.2 \): \( n = \left( \frac{1.96^2 \times 0.2 \times 0.8}{0.02^2} \right) \).
09

Solve the Equation for Sample Size

Calculate \( n = \left( \frac{1.96^2 \times 0.2 \times 0.8}{0.02^2} \right) = 1536.64 \). Round up to 1537, as the sample size must be a whole number.

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Key Concepts

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

Sample Proportion
The sample proportion is a way to estimate a percentage based on a smaller, manageable number of observations. In essence, it provides an idea of what we might expect in the entire population from studying a smaller group. In our example, we have a sample of 200 students, out of which 40 are smokers. To find the sample proportion of smokers, we divide the number of smokers by the total number of students:- Formula: \( \hat{p} = \frac{\text{number of smokers}}{\text{total sample size}} \)Using this, we get \( \hat{p} = \frac{40}{200} = 0.2 \). This indicates that 20% of the sampled students smoke. This proportion (represented as \( \hat{p} \)) serves as a basis for calculating the confidence interval, helping to extrapolate these findings to the broader university student population.
Margin of Error
The margin of error quantifies the range within which the true population parameter is expected to lie. It is a crucial component of confidence interval estimation because it provides the "give or take" fluctuation we expect around our sample proportion. In simple terms, it's like saying we're almost certain our measurement is correct, give or take a little variation.This variation originates from the random sample variability and is strongly influenced by the size of the sample. A larger sample size reduces the margin of error, while a smaller sample increases it.- Formula: \( ME = Z \times SE \)In our exercise, the confidence interval for the proportion of smokers is calculated at 95%, leading to a margin of error derived from a standard Z-value and standard error. This exercise aims to keep the margin of error low, at a 2% threshold, by adjusting the necessary sample size.
Z-value
The Z-value, in the context of statistics, corresponds to the number of standard deviations a data point is from the mean of a distribution. It is used in confidence interval estimation to find how many standard deviations away we need to be to capture a certain percentage of the data distribution. For a 95% confidence interval, which is common in statistical analysis, the Z-value is 1.96. This value tells us that we should move 1.96 standard deviations away from the mean in both directions to encompass 95% of the possible data points. It functions as a multiplier that helps define the range of uncertainty in a confidence interval calculation. - Default Z-value for 95% confidence: 1.96 Using it alongside the standard error allows for constructing the confidence interval, reflecting the assumed reliability of our sample findings as an estimate of the population.
Standard Error
The standard error (SE) measures how accurately a sample statistic estimates a population parameter. It reflects the sampling distribution's standard deviation, which informs how much this distribution varies. The standard error decreases as the sample size increases, indicating more accurate estimation of the population parameter.- Formula for standard error of a proportion: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)In our example, the standard error of the sample proportion is calculated as follows: given \( \hat{p} = 0.2 \) or 20%, and \( n = 200 \), \( SE = \sqrt{\frac{0.2 \times 0.8}{200}} \approx 0.0283 \). This SE value quantifies the average variability we might expect between the sample proportion and the true population proportion if we were to take many samples of the same size from the population.A smaller standard error suggests our sample proportion provides a more accurate estimate of the true population proportion, which is why it's crucial in designing surveys and interpreting data results.

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Most popular questions from this chapter

A professor using an open source introductory statistics book predicts that \(60 \%\) of the students will purchase a hard copy of the book, \(25 \%\) will print it out from the web, and \(15 \%\) will read it online. At the end of the semester he asks his students to complete a survey where they indicate what format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30 said they printed it out from the web, and 25 said they read it online. (a) State the hypotheses for testing if the professor's predictions were inaccurate. (b) How many students did the professor expect to buy the book, print the book, and read the book exclusively online? (c) This is an appropriate setting for a chi-square test. List the conditions required for a test and verify they are satisfied. (d) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value. (e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpret your conclusion in this context.

The Stanford University Heart Transplant Study was conducted to determine whether an experimental heart transplant program increased lifespan. Each patient entering the program was officially designated a heart transplant candidate, meaning that he was gravely ill and might benefit from a new heart. Patients were randomly assigned into treatment and control groups. Patients in the treatment group received a transplant, and those in the control group did not. The table below displays how many patients survived and died in each group. $$ \begin{array}{ccc} \hline & \text { control } & \text { treatment } \\ \hline \text { alive } & 4 & 24 \\ \text { dead } & 30 & 45 \\ \hline \end{array} $$ Suppose we are interested in estimating the difference in survival rate between the control and treatment groups using a confidence interval. Explain why we cannot construct such an interval using the normal approximation. What might go wrong if we constructed the confidence interval despite this problem?

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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.

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