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Just one serving a month of kale or collard greens or more than two servings of carrots a week can reduce the risk of glaucoma by more than \(60 \%,\) according to a UCLA study of 1000 women. Using a \(90 \%\) confidence interval for the true binomial proportion based on this random sample of 1000 binomial trials and an observed proportion of \(0.60,\) estimate the proportion of glaucoma risk reduction in women who eat the recommended servings of kale, collard greens, or carrots.

Short Answer

Expert verified
This estimated proportion of glaucoma risk reduction for women who eat the recommended servings of kale, collard greens, or carrots is within the calculated confidence interval.

Step by step solution

01

Identify the Given Information

We are given that the sample size \(n = 1000\), the observed success proportion \(p = 0.60\), and the confidence level is \(90\%\).
02

Calculate the Standard Error

Standard error, SE, is given by the formula: SE = \(\sqrt{\(p(1 - p) / n}\) Here, \(p = 0.60\) is the observed success proportion and \(n = 1000\) is the sample size. So, substitute these values into the formula to calculate SE.
03

Calculate the Confidence Interval

Confidence interval is given by the formula: CI = \(p ± Z*SE\) Here, \(p = 0.60\) is the observed success proportion, and SE is the standard error calculated in Step 2. \(Z\) is the Z-score, which is based on the confidence level. For a 90% confidence level, \(Z = 1.645\). Substitute these values into the formula to calculate the confidence interval.
04

Interpret the Result

The confidence interval calculated in step 3 gives us the range within which we are \(90 \%\) confident that the true proportion of women for whom eating the recommended servings of the listed vegetables reduces the risk of glaucoma lies.

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Key Concepts

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

Binomial Proportion
When studying the relationship between diet and glaucoma risk, researchers often use a statistical measure called the binomial proportion. This measure refers to the percentage of a specific outcome occurring in a binomial distribution, where there are only two possible outcomes: 'success' or 'failure.'

In the context of the given exercise, a 'success' would be the case where intake of certain vegetables is associated with a reduction of glaucoma risk, and the 'observed proportion' of 0.60 indicates 60% success rate in the sample. It is this proportion that we want to estimate for the entire population, with a certain degree of certainty. When similar educational exercises teach about binomial proportion, they allow students to understand the real-world implications of statistical results in health sciences and other fields.
Statistics Education
Statistics education is crucial, as it arms students with the tools to understand data and make informed decisions based on numerical evidence. In our kale and glaucoma study example, understanding the concept of confidence intervals allows one to comprehend that statistical findings often come with a margin of error.

Effective teaching methods not only include step-by-step solutions but also contextual explanations, practical examples, like this diet study, and visualization techniques. By emphasizing the process of statistical analysis, students can appreciate the significance of statistical methods in research and daily life. Exploring these statistics concepts through engaging exercises supports the development of critical thinking and data literacy skills.
Sample Size
Sample size, denoted as 'n' in statistical notation, plays a critical role in determining the accuracy of a confidence interval. In the provided example, a relatively large sample size of 1000 women ensures that the estimated proportion is likely to be a good representation of the true proportion in the population.

A large sample size tends to yield a smaller standard error, which contributes to a narrower confidence interval, implying greater precision in the estimate. It's essential for students to understand how sample size affects statistical calculations and the quality of the conclusions drawn from data analysis.
Standard Error
Standard error measures the variability or dispersion of sample statistics like the mean or proportion. In the context of confidence intervals, the standard error quantifies how far we would expect the sample estimate to deviate from the true population parameter if we were to take many samples.

In the exercise, calculating the standard error is a pivotal step in developing the confidence interval for the risk reduction proportion. A smaller standard error indicates a more precise estimate, enhancing the reliability of statistical conclusions. Grasping the role of standard error is fundamental for students to understand confidence intervals and their interpretation in research findings.

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Most popular questions from this chapter

"You say tomato, burger lovers say ketchup!" According to a recent T.G.I. Friday's restaurants' random survey of 1027 Americans, approximately half \((47 \%)\) said that ketchup is their preferred burger condiment. The survey quoted a margin of error of plus or minus \(3.1 \% .\) a. Describe how this survey of 1027 Americans fits the properties of a binomial experiment. Specifically identify \(n,\) a trial, success, \(p,\) and \(x\). b. What is the point estimate for the proportion of all Americans who prefer ketchup on their burger? Is it a parameter or a statistic? c. Calculate the \(95 \%\) confidence maximum error of estimate for a binomial experiment of 1027 trials that results in an observed proportion of 0.47 d. How is the maximum error, found in part c, related to the \(3.1 \%\) margin of error quoted in the survey report? e. Find the \(95 \%\) confidence interval for the true proportion \(p\) based on a binomial experiment of 1027 trials that results in an observed proportion of 0.47.

a. What is the relationship between \(p=P(\text { success })\) and \(q=P(\text { failure }) ?\) Explain. b. Explain why the relationship between \(p\) and \(q\) can be expressed by the formula \(q=1-p\) c. If \(p=0.6,\) what is the value of \(q ?\) d. If the value of \(q^{\prime}=0.273,\) what is the value of \(p^{\prime} ?\)

The Pizza Shack in Exercise 9.177 has completed its sampling and the results are in! On Tuesday afternoon, they sampled 15 customers and 9 preferred the new pizza crust. On Friday evening, they sampled 200 customers and 120 preferred the new pizza crust. Help the manager interpret the meaning of these results. Use a one-tailed test with \(H_{a}: p>0.50\) and \(\alpha=0.02 .\) Use \(z\) as the test statistic. a. Is there sufficient evidence to conclude a significant preference for the new crust based on Tuesday's customers? b. Is there sufficient evidence to conclude a significant preference for the new crust based on Friday's customers? c. since the percentage of customers preferring the new crust was the same, \(p^{\prime}=0.60\) in both samplings, explain why the answers in parts a and b are not the same.

Three nationwide poll results are described below. USA Today Snapshot/Rent.com, August 18,2009 \(N=1000\) adults 18 and over; \(\mathrm{MoE} \pm 3 .\) (MoE is margin of error. "What renters look for the most when seeking an apartment:" Washer/dryer\(-39\%,\) Air Conditioning \(-30 \%,\) Fitness Center- \(10 \%,\) Pool \(-10 \%\) USA Today/Harris Interactive Poll, February \(10-15,2009 ; N=1010\) adults; MoE ±3. "Americans who say people on Wall Street are "as honest and moral as other people." Disagree \(-70 \%\) Agree \(-26 \%,\) Not sure/refuse to answer \(-4 \%\) American Association of Retired Persons Bulletin/AARP survey, July 22-August 2, 2009; \(N=1006\) adults age 50 and older; \(\mathrm{MoE} \pm 3\). The American Association of Retired Persons Bulletin Survey reported that \(16 \%\) of adults, 50 and older, said they are likely to return to school. Each of the polls is based on approximately 1005 randomly selected adults. a. Calculate the \(95 \%\) confidence maximum error of estimate for the true binomial proportion based on binomial experiments with the same sample size and observed proportion as listed first in each article. b. Explain what caused the values of the maximum errors to vary. c. The margin of error being reported is typically the value of the maximum error rounded to the next larger whole percentage. Do your results in part a verify this? d. Explain why the round-up practice is considered "conservative." e. What value of \(p\) should be used to calculate the standard error if the most conservative margin of error is desired?

Use a computer or calculator to find the \(p\) -value for the following hypothesis test: \(H_{o}: \sigma=12.4\) versus \(H_{a}: \sigma>12.4,\) if \(\chi^{2} \star=36.59\) for a sample of \(n=24\).

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