/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Campus Life: Coeds What percenta... [FREE SOLUTION] | 91Ó°ÊÓ

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Campus Life: Coeds What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p\), how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) The Statistical Abstract of the United States, 112 th Edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

Short Answer

Expert verified
(a) 664; (b) 660.

Step by step solution

01

Understanding the Confidence Interval Formula

To solve both parts of this problem, we need the formula for determining sample size for a proportion within a specified margin of error at a given confidence level. The formula is: \[ n = \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \]where \(n\) is the sample size, \(Z\) is the Z-value from the standard normal distribution for a 99% confidence level (which is 2.576), \(E\) is the margin of error (0.05), and \(p\) is the estimated proportion of the population (proportion of female students in this case).
02

Calculating Sample Size Without Preliminary Estimate

For part (a), when no preliminary estimate of \(p\) is available, we assume the most conservative estimate, \(p = 0.5\). This is because the product \(p(1-p)\) achieves its maximum value at \(p=0.5\), which ensures the largest possible sample size and hence generalizes across the widest range of \(p\) values.Substituting these values into our formula: \[ n = \frac{(2.576)^2 \cdot 0.5 \cdot (1-0.5)}{0.05^2} \] \[ n = \frac{6.635776 \cdot 0.25}{0.0025} \] \[ n = \frac{1.658944}{0.0025} \approx 663.58 \] Round up, since you can't survey a fraction of a student. Therefore, \(n = 664\).
03

Calculating Sample Size Using Preliminary Estimate

For part (b), use the estimate \(p = 0.54\) from the Statistical Abstract. Substitute this value into the formula:\[ n = \frac{(2.576)^2 \cdot 0.54 \cdot (1-0.54)}{0.05^2} \] \[ n = \frac{6.635776 \cdot 0.2484}{0.0025} \] \[ n = \frac{1.648234\}{0.0025} \approx 659.29 \] Round up to ensure a sufficiently large sample size. Therefore, \(n = 660\).

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Key Concepts

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

Confidence Interval
Understanding the concept of a confidence interval is crucial in statistics, especially when trying to make inferences about a population based on sample data. A confidence interval provides a range of plausible values for a population parameter, such as a proportion, based on the sample data collected. Typically, this range is constructed around a point estimate, such as a sample mean or proportion, and incorporates the margin of error to reflect the uncertainty inherent in the sampling process.
The confidence level, usually expressed as a percentage (for example, 99%), indicates the probability that the confidence interval contains the true population parameter. Higher confidence levels result in wider confidence intervals, reflecting greater certainty about capturing the true parameter. In situations involving population proportions, a confidence interval for the proportion, denoted as \(\hat{p}\), is calculated using the formula:
  • Confidence Interval = \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
Where \(Z\) is the Z-score corresponding to the desired confidence level, \(\hat{p}\) is the sample proportion, and \(n\) is the sample size.
Understanding this concept lays the groundwork for accurate statistical analysis and decision-making in various fields.
Sample Size Calculation
Calculating a sample size is a fundamental step in designing a study or survey to ensure the results are reliable and valid. The sample size influences the precision of the estimates and the width of the confidence interval. A larger sample size reduces the margin of error, providing a more precise estimate of the population parameter.
To calculate the necessary sample size for estimating a population proportion with a specific margin of error at a certain confidence level, the following formula is used:
  • Sample Size, \ = \frac{Z^2 \, p \, (1 - p)}{E^2}
Where \(n\) is the sample size, \(Z\) is the Z-value from the standard normal distribution reflecting the desired confidence level, \(p\) is the estimated population proportion, and \(E\) is the desired margin of error. For example, in a scenario with a 99% confidence level and a margin of error of 0.05, the Z-value would be 2.576.
Choosing \(p\) when it's unknown, the safest assumption is to use \(p = 0.5\) since it provides the maximum variability, hence the largest required sample size. Adjustments to \(p\) with prior estimates can lead to smaller required sample sizes, improving feasibility and cost-effectiveness of conducting the survey or study.
Proportion Estimation
Proportion estimation in statistics is about estimating the proportion of a certain attribute in a population based on a sample. This is common in surveys, opinion polls, and market research. Estimating a population proportion involves gathering data, calculating the sample proportion, and then using statistical methods to infer the population proportion.
Using samples to estimate proportions is crucial because it is often impractical or impossible to measure an entire population. Instead, a representative sample is chosen to estimate the true proportion, denoted as \(p\), of a particular characteristic within the population. From this sample, we calculate the sample proportion, \(\hat{p}\), by dividing the number of times the characteristic of interest is observed by the total sample size.
The accuracy of this estimation is enhanced by employing confidence intervals, which provide a range where the true proportion is expected to lie. Proportion estimation becomes all the more reliable with appropriate sample size, ensuring that the estimates are close to the true population values with an acceptable amount of variation, or margin of error. By refining these elements of statistics education, students are better equipped to conduct credible research and analysis.

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Most popular questions from this chapter

Focus Problem: Wood Duck Nests In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 474 eggs in group I boxes, of which a field count showed about 270 had hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 805 eggs in group II boxes, of which a field count showed about 270 had hatched. (a) Find a point estimate \(\hat{p}_{1}\) for \(p_{1}\), the proportion of eggs that hatched in group I nest box placements. Find a \(95 \%\) confidence interval for \(p_{1}\). (b) Find a point estimate \(\hat{p}_{2}\) for \(p_{2}\), the proportion of eggs that hatched in group II nest box placements. Find a \(95 \%\) confidence interval for \(p_{2}\). (c) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\). Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes? (d) Interpretation What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

Psychology: Self-Esteem Female undergraduates in randomized groups of 15 took part in a self-esteem study ("There's More to Self-Esteem than Whether It Is High or Low: The Importance of Stability of Self-Esteem," by M. H. Kernis et al., Journal of Personality and Social Psychology, Vol. 65, No. 6). The study measured an index of self-esteem from the points of view competence, social acceptance, and physical attractiveness. Let \(x_{1}, x_{2}\), and \(x_{3}\) be random variables representing the measure of self-esteem through \(x_{1}\) (competence), \(x_{2}\) (social acceptance), and \(x_{3}\) (attractiveness). Higher index values mean a more positive influence on self-esteem. $$ \begin{array}{ccccc} \hline \text { Variable } & \text { Sample Size } & \text { Mean } \bar{x} & \text { Standard Deviation } s & \text { Population Mean } \\ \hline x_{1} & 15 & 19.84 & 3.07 & \mu_{1} \\ x_{2} & 15 & 19.32 & 3.62 & \mu_{2} \\ x_{3} & 15 & 17.88 & 3.74 & \mu_{3} \\ \hline \end{array} $$ (a) Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (b) Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{3} .\) (c) Find an \(85 \%\) confidence interval for \(\mu_{2}-\mu_{3}\). (d) Interpretation Comment on the meaning of each of the confidence intervals found in parts (a), (b), and (c). At the \(85 \%\) confidence level, what can you say about the average differences in influence on self-esteem between competence and social acceptance? between competence and attractiveness? between social acceptance and attractiveness?

If a \(90 \%\) confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between \(p_{1}\) and \(p_{2}\) at the \(90 \%\) confidence level?

Basic Computation: Confidence Interval for \(p\) Consider \(n=100\) binomial trials with \(r=30\) successes. (a) Check Requirements Is it appropriate to use a normal distribution to approximate the \(\hat{p}\) distribution? (b) Find a \(90 \%\) confidence interval for the population proportion of successes \(p .\) (c) Interpretation Explain the meaning of the confidence interval you computed.

Psychology: Parental Sensitivity "Parental Sensitivity to Infant Cues: Similarities and Differences Between Mothers and Fathers" by M. V. Graham (Journal of Pediatric Nursing, Vol. 8, No. 6 ) reports a study of parental empathy for sensitivity cues and baby temperament (higher scores mean more empathy). Let \(x_{1}\) be a random variable that represents the score of a mother on an empathy test (as regards her baby). Let \(x_{2}\) be the empathy score of a father. A random sample of 32 mothers gave a sample mean of \(\bar{x}_{1}=69.44\). Another random sample of 32 fathers gave \(\bar{x}_{2}=59 .\) Assume that \(\sigma_{1}=11.69\) and \(\sigma_{2}=11.60\). (a) Check Requirements Which distribution, normal or Student's \(t\), do we use to approximate the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? Explain. (b) Let \(\mu_{1}\) be the population mean of \(x_{1}\) and let \(\mu_{2}\) be the population mean of \(x_{2}\). Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you about the relationship between average empathy scores for mothers compared with those for fathers at the \(99 \%\) confidence level?

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