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Appropriate use of the interval $$ \hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ requires a large sample. For each of the following combinations of \(n\) and \(\hat{p}\), indicate whether the sample size is large enough for this interval to be appropriate. a. \(n=100\) and \(\hat{p}=0.70\) b. \(n=40\) and \(\hat{p}=0.25\) c. \(n=60\) and \(\hat{p}=0.25\) d. \(n=80\) and \(\hat{p}=0.10\)

Short Answer

Expert verified
a. Yes, the sample size is large enough. b. Yes, the sample size is large enough. c. Yes, the sample size is large enough. d. No, the sample size is not large enough.

Step by step solution

01

Case A: n = 100, p-hat = 0.70

First, calculate \(n\hat{p}\) and \(n(1 - \hat{p})\): \(n\hat{p} = 100 \cdot 0.7 = 70\) \(n(1 - \hat{p}) = 100 \cdot 0.3 = 30\) Since both values are greater than or equal to 10, the sample size is large enough for the interval to be appropriate.
02

Case B: n = 40, p-hat = 0.25

Now, let's calculate \(n\hat{p}\) and \(n(1 - \hat{p})\) for this case: \(n\hat{p} = 40 \cdot 0.25 = 10\) \(n(1 - \hat{p}) = 40 \cdot 0.75 = 30\) Both values are greater than or equal to 10, so the sample size is large enough for the interval to be appropriate.
03

Case C: n = 60, p-hat = 0.25

Now, let's calculate \(n\hat{p}\) and \(n(1 - \hat{p})\) for the third case: \(n\hat{p} = 60 \cdot 0.25 = 15\) \(n(1 - \hat{p}) = 60 \cdot 0.75 = 45\) Since both values are greater than or equal to 10, the sample size is large enough for the interval to be appropriate.
04

Case D: n = 80, p-hat = 0.10

Finally, let's calculate \(n\hat{p}\) and \(n(1 - \hat{p})\) for the last case: \(n\hat{p} = 80 \cdot 0.1 = 8\) \(n(1 - \hat{p}) = 80 \cdot 0.9 = 72\) In this case, \(n\hat{p} < 10\), so the sample size is not large enough for the interval to be appropriate.

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Key Concepts

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

Confidence Interval
A confidence interval (CI) is, essentially, a range of values used to estimate the true value of a population parameter. When statisticians mention a '95% confidence interval', it means that if we were to take 100 different samples and compute a CI for each sample, we would expect the true population parameter to fall within those intervals 95 times.

To construct a confidence interval for a proportion, the formula \(\hat{p} \pm(z \text{ critical value}) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) is used, where \(\hat{p}\) is the sample proportion, \(z\) is the z-score corresponding to the desired confidence level, and \(n\) is the sample size. The CI gives a range within which the true population proportion is likely to be found. However, it's validity is based on the assumption that the sample size is sufficiently large and the population distribution is approximately normal.
Sample Size Determination
Determining the appropriate sample size for a study is crucial for accurate estimations. A sample size that's too small may not adequately represent the population, leading to inaccurate estimates and a wider confidence interval. Conversely, an unnecessarily large sample size may be a waste of resources.

The essential criteria for sample size determination often require that both \(n\hat{p}\) and \(n(1-\hat{p})\) should be greater than or equal to 10 for the normal approximation to be valid. If these products fall below 10, the sample is likely too small to yield reliable interval estimates. Moreover, larger samples reduce the margin of error, leading to a more precise confidence interval.
Normal Approximation
The normal approximation is a statistical technique used to approximate the distribution of various statistics, including proportions. When the sample size is large, the distribution of the sample proportion (\(\hat{p}\)) tends to be normally distributed, thanks to the Central Limit Theorem.

This approximation allows us to use the z-score, which is standard in a normal distribution, to calculate our confidence intervals for proportions. However, this depends on having a sufficiently large sample size. Ideally, having \(n\hat{p}\) and \(n(1 - \hat{p})\) greater than or equal to 10 signifies that the normal approximation can be safely used.
Proportion Estimation
Estimating proportions involves finding a point estimate, such as \(\hat{p}\), and a confidence interval to indicate the precision of the estimate. When a proportion is estimated from a sample, statisticians use the formula mentioned earlier to calculate the confidence interval. This estimation, however, is accurate only if the sample used is representative of the population.

For the estimation to be reliable, it's necessary that the sample meets the size requirement for normal approximation. Otherwise, the estimate might not account for the variability within the population correctly, leading to an invalid confidence interval.

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Most popular questions from this chapter

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are registered to vote. For which of the following situations will the estimate tend to be closest to the actual value of \(p ?\) I. \(\quad n=1000, p=0.5\) II. \(\quad n=200, p=0.6\) III. \(n=100, p=0.7\)

In a survey of 800 college students in the United States, 576 indicated that they believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action ("Listening to Dissenting Views Part of Civil Debate," USA TODAY, November 17,2015 ). Assuming that the sample is representative of college students in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of college students who have this belief.

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A researcher wants to estimate the proportion of students enrolled at a university who are registered to vote. Would the standard error of the sample proportion \(\hat{p}\) be larger if the actual population proportion was \(p=0.4\) or \(p=0.8 ?\)

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