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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

The report "The 2016 Consumer Financial Literacy Survey" (The National Foundation for Credit Counseling, www.nfcc.org, retrieved October 28,2016 ) summarized data from a representative sample of 1668 adult Americans. Based on data from this sample, it was reported that over half of U.S. adults would give themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\) on their knowledge of personal finance. This statement was based on observing that 934 people in the sample would have given themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\). a. Construct and interpret a \(95 \%\) confidence interval for the proportion of all adult Americans who would give themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\) on their financial knowledge of personal finance. b. Is the confidence interval from Part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of \(\mathrm{A}\) or \(\mathrm{B}\) ? Explain why or why not.

In spite of the potential safety hazards, some people would like to have an Internet connection in their car. A preliminary survey of adult Americans has estimated the proportion of adult Americans who would like Internet access in their car to be somewhere around 0.30 (USA TODAY, May 1 , 2009). Use the given preliminary estimate to determine the sample size required to estimate this proportion with a margin of error of 0.02

The paper "Sleeping with Technology: Cognitive, Affective and Technology Usage Predictors of Sleep Problems Among College Students" (Sleep Health [2016]: 49-56) summarized data from a survey of a sample of college students. Of the 734 students surveyed, 125 reported that they sleep with their cell phones near the bed and check their phones for something other than the time at least twice during the night. For purposes of this exercise, assume that this sample is representative of college students in the United States. a. Use the given information to estimate the proportion of college students who check their cell phones for something other than the time at least twice during the night. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

Describe how each of the following factors affects the width of the large- sample confidence interval for \(p\) : a. The confidence level b. The sample size c. The value of \(\hat{p}\)

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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