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Do you like tofu? You randomly sample five students at your school to estimate the proportion of students who like tofu. All five students say they like it. a. Find the sample proportion who like it. b. Find the standard error. Does its usual interpretation make sense? c. Find a \(95 \%\) confidence interval, using the large-sample formula. Is it sensible to conclude that all students at your school like tofu? d. Why is it not appropriate to use the ordinary largesample confidence interval in part \(\mathrm{c}\) ? Use a more appropriate approach, and interpret the result.

Short Answer

Expert verified
The appropriate confidence interval is [0.52, 1.00], indicating that between 52% and 100% of students may like tofu.

Step by step solution

01

Calculate Sample Proportion

To find the sample proportion \(\hat{p}\), you divide the number of students who like tofu (which is 5) by the total number of students sampled (which is also 5). Thus, \[\hat{p} = \frac{5}{5} = 1\] This means that 100% of the sampled students like tofu.
02

Calculate Standard Error

The standard error (SE) of a sample proportion is given by the formula \[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Substituting in the values, \[SE = \sqrt{\frac{1 \times (1-1)}{5}} = 0\] Because the sample proportion is 1, the standard error is 0, which might not offer practical interpretation since it assumes no variability.
03

Calculate 95% Confidence Interval (Large-sample Formula)

The formula for a 95% confidence interval for a proportion is\[\hat{p} \pm z \cdot SE\]where \(z\) is the z-score corresponding to a 95% confidence level (approximately 1.96). Since the SE is 0,\[1 \pm 1.96(0) = 1\].The confidence interval is [1,1], which incorrectly suggests all students like tofu.
04

Explanation of Large-sample Formula Error

The large-sample confidence interval assumes that the sample size is large enough for the normal approximation of the binomial distribution to be valid, which is not the case here (total sample size is 5). This results in an unreliable confidence interval that doesn't actually reflect the true variability in the student population.
05

Use More Appropriate Approach (Adjusted Wald Interval)

The Adjusted Wald, or Agresti-Coull interval, is better for small samples. First, you adjust the sample counts: \[\tilde{p} = \frac{x+z^2/2}{n+z^2}\]Here, \(x = 5, z^2/2 = 1.96^2/2 \approx 1.92\)\[\tilde{p} = \frac{5+1.92}{5+1.96^2} \approx 0.78\]Then find the interval:\[0.78 \pm 1.96 \times \sqrt{\frac{0.78(1-0.78)}{10.84}} = [0.52, 1.00]\].This suggests between 52% and 100% of students at school like tofu.

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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 statistical measure that gives us a view into how a subset from a larger group behaves. In simple terms, it's about finding out what proportion of your sampled group agrees with a certain statement or trait. For instance, to understand the liking for tofu, you take a small group from a larger pool—in this case, five students from the school—and ask them if they liked tofu.
  • To find the sample proportion, you divide the number of favorable responses by the total number of responses. If all five students say they like tofu, the sample proportion is: \[ \hat{p} = \frac{5}{5} = 1 \]
  • This result means that 100% of the sampled students like tofu.
This might make you think everyone at the school likes tofu, but it's just a proportion of your sample, not the entire population.
Standard Error
The standard error (SE) is an essential concept in statistics as it measures the variability or spread of a sample statistic. It helps us understand how much the sample proportion might differ from the actual population proportion if you were to take multiple samples. For the sample proportion calculation, the standard error formula is: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \] In our tofu exercise, if everyone in your tiny sample group likes tofu, then \[ \hat{p} = 1. \]This leads the standard error to be zero, calculated as: \[ SE = \sqrt{ \frac{1 \times (1-1)}{5} } = 0 \] This value of zero could be misleading because it suggests there's no variability; however, in practice, it implies a lack of information about how the whole population might respond, as all the sampled individuals gave the same answer. Standard error values are more meaningful with more varied data and larger samples.
Confidence Interval
Confidence Intervals (CI) provide a range that is likely to contain the true population proportion. It is a key concept used to indicate the reliability of an estimate. Usually, a 95% confidence interval is selected to suggest that if you were to sample many times, 95% of the intervals would contain the actual population proportion. The formula for calculating a 95% confidence interval for a sample proportion typically looks like: \[ \hat{p} \pm z \cdot SE \]Where "\(z\)" represents the z-score that corresponds to your chosen confidence level (approximately \(1.96\) for a 95% confidence level). With our zero standard error from above, this means:\[ 1 \pm 1.96 \times 0 = [1, 1] \]While this narrow interval suggests certainty that you're correct about everyone liking tofu, it's derived from the flawed assumption that all students are like the sampled ones, which is not always sensible. Especially with small samples, this assumption can lead to a misleading overconfidence in the results.
Binomial Distribution
The binomial distribution is fundamental in statistics, particularly concerning binary data—where outcomes are "success" or "failure". It models the number of successful outcomes—like liking tofu—in a fixed number of trials, such as the number of students asked.
  • In our tofu scenario, each trial (each student asked) can be seen as a success (student likes tofu) or a failure (student does not like tofu). The binomial distribution would model these outcomes.
  • The large-sample confidence interval assumes normal approximation should work, but for small samples, like the five students, the binomial nature can greatly affect outcome reliability.
When working with small sample sizes, approaches like the Adjusted Wald Interval need to be utilized because they offer better accuracy. They ensure that the interval accounts for variability you might see if more samples are taken, preserving the integrity of the statistical inference.

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Most popular questions from this chapter

When the 2000 GSS asked subjects (variable GRNSOL) if they would be willing to accept cuts in their standard of living to protect the environment, 344 of 1170 subjects said yes. a. Estimate the population proportion who would answer yes. b. Find the margin of error for a \(95 \%\) confidence interval for this estimate. c. Find a \(95 \%\) confidence interval for that proportion. What do the numbers in this interval represent? d. State and check the assumptions needed for the interval in part \(c\) to be valid.

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 $$\$ 509$$

Multiple choice: Why z? The reason we use a z-score from a normal distribution in constructing a large-sample confidence interval for a proportion is that a. For large random samples the sampling distribution of the sample proportion is approximately normal. b. The population distribution is normal. c. For large random samples the data distribution is approximately normal. d. For any \(n\) we use the \(t\) distribution to get a confidence interval, and for large \(n\) the \(t\) distribution looks like the standard normal distribution.

Wife doesn't want kids The 1996 GSS asked, "If the husband in a family wants children, but the wife decides that she does not want any children, is it all right for the wife to refuse to have children?" Of 699 respondents, 576 said yes. a. Find a \(99 \%\) confidence interval for the population proportion who would say yes. Can you conclude that the population proportion exceeds \(75 \%\) ? Why? b. Without doing any calculation, explain whether the interval in part a would be wider or narrower than a \(95 \%\) confidence interval for the population proportion who would say yes.

Psychologists' income In \(2003,\) the American Psychological Association conducted a survey (at research.apa.org) of a random sample of psychologists to estimate mean incomes for psychologists with various academic degrees and levels of experience. Of the 31 psychologists who received a masters degree in \(2003,\) the mean income was \(\$ 43,834\) with a standard deviation of \(\$ 16,870\) a. Construct a \(95 \%\) confidence interval for the population mean. Interpret. b. What assumption about the population distribution of psychologists' incomes does the confidence interval method make? c. If the assumption about the shape of the population distribution is not valid, does this invalidate the results? Explain.

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