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Chapter 8: Problem 56

Do students like statistics? All respondents out of a random sample of ten students in a college said that they like statistics. Now you want to estimate the proportion of students who like statistics in the whole college. a. Find the sample proportion of students who like statistics. b. Find the standard error of the estimate and interpret it. c. Find a \(95 \%\) confidence interval using the large-sample formula. Is it appropriate to use the ordinary largesample confidence interval to obtain an estimation for the population proportion? d. Why is it not appropriate to use the ordinary largesample confidence interval in part c? Use a more appropriate approach and interpret the result.

Short Answer

Expert verified
a. Sample proportion = 1.0; b. SE = 0; c. The regular interval is '1.0 to 1.0', inappropriate due to no variability; d. Use Wilson Score Interval for a better estimate.

Step by step solution

01

Calculate the Sample Proportion

The sample proportion (\( \hat{p} \)) is calculated by dividing the number of students in the sample who like statistics by the total number of students in the sample. Here, all 10 students like statistics, so the sample proportion is \( \hat{p} = \frac{10}{10} = 1.0 \).
02

Calculate the Standard Error of the Estimate

The standard error (SE) for the sample proportion is calculated using the formula: \[ \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \( n \) is the sample size. Plugging in the values: \( \hat{p} = 1.0 \) and \( n = 10 \), we get: \[ \text{SE} = \sqrt{\frac{1.0 \times (1-1.0)}{10}} = \sqrt{0} = 0 \]. Interpret this as having no variability in the estimate due to the sample showing unanimous agreement.
03

Assess Large-Sample Confidence Interval Suitability

To calculate a 95% confidence interval for the population proportion, we usually use: \[ \hat{p} \pm Z \times \text{SE} \] where \( Z \) is the Z-score for 95% confidence (approximately 1.96). However, because \( \text{SE} = 0 \), this method suggests an interval of '1.0 to 1.0', which is overly precise and derived from a non-diverse sample. Thus, use this method inappropriately for proportions near 0 or 1.
04

Use a More Appropriate Confidence Interval Approach

In cases where the sample proportion is at the extremes (like 0 or 1), use the Wilson Score Interval or Adjusted Wald Interval for a more accurate confidence interval. For the Wilson Score Interval: \[ \text{Adjusted interval is found with: } \hat{p}_w = \frac{x + 1.96^2/2}{n + 1.96^2} \] \( n_{w} = n + 1.96^2 \). Calculate: \( \hat{p}_w = \frac{10 + 1.96^2/2}{10 + 1.96^2} \approx 0.96 \), and the 95% interval: \[ CI = \hat{p}_w \pm 1.96 \sqrt{\frac{\hat{p}_w (1-\hat{p}_w)}{n_{w}}} \].The Wilson Score gives a more cautious estimate, reflecting practical reporting of student preferences rather than assuming complete certainty.
05

Explain Why Ordinary Confidence Interval is Inappropriate

The ordinary confidence interval assumes variation in the sample, typically fitted by the normal distribution. For a proportion of 1.0, this assumption is violated leading to the assertion that the true proportion might be exactly 1.0, which lacks a realistic basis considering potential diversity in larger samples.

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Key Concepts

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

Confidence Intervals
Confidence intervals are a vital concept in statistical inference. They provide a range of values which likely include the true population parameter. The idea is to offer a range of plausible values for the population characteristic based on sample data.

To calculate a confidence interval, especially for proportions, we often use the formula:
  • \( \hat{p} \pm Z \times \text{SE} \)
where \( \hat{p} \) is the sample proportion, \( Z \) is the Z-score corresponding to the desired confidence level, and \( \text{SE} \) is the standard error.

A 95% confidence interval is widely used. The Z-score for a 95% confidence level is approximately 1.96. However, when a sample proportion is extreme (like 0 or 1), this standard approach can yield misleading results, making specialized methods such as the Wilson Score Interval more appropriate. This alternative provides a more conservative and realistic estimate by adjusting for the possibility of diversity absent in small samples.
Sample Proportion
The sample proportion is a straightforward yet powerful statistic in understanding a population's characteristics based on sample data. It represents the fraction of the sample that exhibits a particular trait or behavior.

You calculate it by dividing the number of favorable outcomes by the total number of observations:
  • \( \hat{p} = \frac{x}{n} \)
where \( x \) is the count of successes (e.g., students who like statistics) and \( n \) is the total sample size.

In the scenario presented, all 10 students reported enjoying statistics, leading to a sample proportion of 1.0. Such an absolute result can be both a strength and a weakness. On the one hand, it strongly suggests a positive preference within the sample. On the other, it raises questions about how well this might reflect a more varied or larger population. Extreme proportions highlight the need for caution in generalizing results.
Standard Error
The standard error (SE) is crucial for understanding the reliability of a sample statistic. It measures the variability or dispersion of the sample proportion around the true population proportion.

The standard error for a sample proportion is calculated as:
  • \( \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
where \( \hat{p} \) is the sample proportion and \( n \) is the sample size.

In cases where the sample proportion is at the extremes (like 0 or 1), the SE becomes zero or very low. This result indicates almost no variability within the sample. However, an SE of 0 is not very informative for larger population predictions. In such scenarios, alternative methods like the Adjusted Wald Interval should be applied. These methods guard against the overconfidence that zero variability can suggest, ensuring more representative confidence intervals that account for potential diversity in the population.

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Most popular questions from this chapter

According to a survey conducted by KPMG in \(2016,\) almost \(46 \%\) of the surveyed companies intend to grow their operations in Luxembourg over the next two years (Source: https://www.kpmg.com/LU/en/IssuesAndInsights/ Articlespublications/Documents/ManagementCompany-CEO-Survey-032016.pdf). a. Can you specify the assumptions made to construct a \(95 \%\) confidence interval for the population proportion? b. If the sample size is 5000 , verify that the assumptions of part a are satisfied and construct the \(95 \%\) confidence interval. Determine whether the proportion of companies who intend to grow their operations in Luxembourg is a majority or a minority.

Wage discrimination? According to a union agreement, the mean income for all senior-level assembly-line workers in a large company equals \(\$ 500\) per week. A representative of a women's group decides to analyze whether the mean income for female employees matches this norm. For a random sample of nine female employees, using software, she obtains a \(95 \%\) confidence interval of ( 371 , 509 ). Explain what is wrong with each of the following interpretations of this interval. a. We infer that \(95 \%\) of the women in the population have income between \(\$ 371\) and \(\$ 509\) per week. b. If random samples of nine women were repeatedly selected, then \(95 \%\) of the time the sample mean income would be between \(\$ 371\) and \(\$ 509\). c. We can be \(95 \%\) confident that \(\bar{x}\) is between \(\$ 371\) and \(\$ 509 .\) d. If we repeatedly sampled the entire population, then \(95 \%\) of the time the population mean would be between \(\$ 371\) and \(\$ 509 .\)

Abstainers The Harvard study mentioned in the previous exercise estimated that \(19 \%\) of college students abstain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.05 with probability \(0.95,\) if before conducting the study a. You are unwilling to guess the proportion value at your school? b. You use the Harvard study as a guideline? c. Use the results from parts a and b to explain why strategy (a) is inefficient if you are quite sure you'll get a sample proportion that is far from 0.50 .

Grandmas using e-mail For the question about e-mail in the previous exercise, the 14 females in the GSS of age at least 80 had the responses $$ 0,0,0,0,1,1,1,2,2,6,6,7,7,10 $$ a. Using the web app, software or a calculator, find the sample mean and standard deviation and the standard error of the sample mean. b. Find and interpret a \(90 \%\) confidence interval for the population mean. c. Explain why the population distribution may be skewed right. If this is the case, is the interval you obtained in part b useless, or is it still valid? Explain.

Effect of \(n\) Find the margin of error for a \(95 \%\) confidence interval for estimating the population mean when the sample standard deviation equals \(100,\) with a sample size of (i) 25 and (ii) 100 . What is the effect of increasing the sample size? (You can use Table \(\mathrm{B}\) in the back to find the appropriate \(t\) -scores.)
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