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91Ó°ÊÓ

In estimating the mean score on a fitness exam, we use an original sample of size \(n=30\) and a bootstrap distribution containing 5000 bootstrap samples to obtain a \(95 \%\) confidence interval of 67 to 73 . In Exercises 3.90 to 3.95 , a change in this process is described. If all else stays the same, which of the following confidence intervals \((A, B,\) or \(C)\) is the most likely result after the change: \(\begin{array}{lll}A .66 \text { to } 74 & B .67 \text { to } 73 & \text { C. } 67.5 \text { to } 72.5\end{array}\) Using an original sample of size \(n=16\)

Short Answer

Expert verified
After the change of using an original sample size of \(n=16\), interval A (66 to 74) is the most likely.

Step by step solution

01

Understand the sample size effect

Remember that when the sample size decreases, the variability tends to increase. This means that our confidence interval will likely become wider.
02

Compare given intervals

Now have a look at the three given intervals \((A, B, and C)\). Interval \(A\) is from 66 to 74 (width = 8), interval \(B\) is from 67 to 73 (width = 6, same as in our original problem), and interval \(C\) is from 67.5 to 72.5 (width = 5). Based on our prior analysis, we're looking for a wider interval.
03

Select most likely interval

Given the need for a wider interval and based on the above mentioned effect of a smaller sample size, interval \(A\) which ranges from 66 to 74 is the most probable interval, since it's wider compared to the others.

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Key Concepts

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

The Impact of Sample Size on Confidence Intervals
Understanding the effect of sample size on the confidence interval is paramount when conducting statistical analysis. A confidence interval provides a range within which we can say, with a certain level of confidence, that the true parameter of the population lies.

When dealing with smaller sample sizes, like the change in the exercise from an original sample of 30 to a sample of 16, increased variability in outcomes is a standard statistical phenomenon. As the sample size diminishes, each data point has a greater impact on the overall estimate, leading to less precise measurements. Consequently, the confidence interval must adjust to reflect this increased uncertainty - typically becoming wider to maintain the same level of confidence.

Therefore, when confronted with interval options after a reduction in sample size, it is logical to predict that the wider interval is more plausible. For example, from the given choices, interval A's width of 8 suggests greater uncertainty, which aligns with the effect of a reduced sample size, making it a likely result.
Understanding Bootstrap Distribution
The bootstrap distribution is a powerful tool in modern statistics, used for estimating the sampling distribution of a statistic by repeatedly sampling with replacement from the original sample data.

In the given exercise, the bootstrap distribution contains 5000 bootstrap samples. By resampling the original data multiple times, each bootstrap sample creates its version of the estimate. These repetitions form a distribution of estimates - the bootstrap distribution. This distribution approximates the variability of the statistic across different possible samples from the population.

Essentially, the bootstrap method allows us to assess the stability of our estimate. It tells us how the mean score on the fitness exam may vary from one sample to another. Therefore, understanding how the bootstrap distribution is formed and interpreted underpins our ability to make reliable inferences about population parameters from sample data.
Statistical Variability and Its Effects
Statistical variability, also known as statistical variation or noise, is the inherent variability in data that occurs due to natural differences across observations. It's a concept that measures how much the data values differ from each other.

High variability in data implies that the sample points are spread out over a wider range, potentially indicating a diverse population. Low variability suggests that the data points are clustered closely around the mean. In the context of confidence intervals, higher statistical variability would usually result in less precision, as it would widen the range of values thought to contain the population parameter.

This variability is a crucial factor in determining the reliability of statistical estimates. For instance, in the problem provided, the key to selecting the most probable confidence interval after a reduction in sample size was recognizing that this change would likely increase the data's variability. The larger interval accommodates this greater variability, ensuring that we still capture the true mean score with the level of confidence we desire.

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Most popular questions from this chapter

Daily Tip Revenue for a Waitress Data 2.12 on page 119 describes information from a sample of 157 restaurant bills collected at the First Crush bistro. The data is available in RestaurantTips. Two intervals are given below for the average tip left at a restaurant; one is a \(90 \%\) confidence interval and one is a \(99 \%\) confidence interval. Interval A: 3.55 to \(4.15 \quad\) Interval B: 3.35 to 4.35 (a) Which one is the \(90 \%\) confidence interval? Which one is the \(99 \%\) confidence interval? (b) One waitress generally waits on 20 tables in an average shift. Give a range for her expected daily tip revenue, using both \(90 \%\) and \(99 \%\) confidence. Interpret your results.

What Is the Effect of Including Some Indifferent Fish? In the experiment described above under Fish Democracies, the schools of fish in the study with an opinionated minority and a less passionate majority picked the majority option only about \(17 \%\) of the time. However, when groups also included 10 fish with no opinion, the schools of fish picked the majority option \(61 \%\) of the time. We want to estimate the effect of adding the fish with no opinion to the group, which means we want to estimate the difference in the two proportions. We learn from the study that the standard error for estimating this difference is about \(0.14 .\) Define the parameter we are estimating, give the best point estimate, and find and interpret a \(95 \%\) confidence interval. Is it plausible that adding indifferent fish really has no effect on the outcome?

Playing Video Games A new study provides some evidence that playing action video games strengthens a person's ability to translate sensory information quickly into accurate decisions. Researchers had 23 male volunteers with an average age of 20 look at moving arrays on a computer screen and indicate the direction in which the dots were moving. \(^{26}\) Half of the volunteers (11 men) reported playing action video games at least five times a week for the previous year, while the other 12 reported no video game playing in the previous year. The response time and the accuracy score were both measured. A \(95 \%\) confidence interval for the mean response time for game players minus the mean response time for non-players is -1.8 to -1.2 seconds, while a \(95 \%\) confidence interval for mean accuracy score for game players minus mean accuracy score for non-players is -4.2 to +5.8 . (a) Interpret the meaning of the \(95 \%\) confidence interval for difference in mean response time. (b) Is it likely that game players and non-game players are basically the same in response time? Why or why not? If not, which group is faster (with a smaller response time)? (c) Interpret the meaning of the \(95 \%\) confidence interval for difference in mean accuracy score. (d) Is it likely that game players and non-game players are basically the same in accuracy? Why or why not? If not, which group is more accurate?

Give information about the proportion of a sample that agrees with a certain statement. Use StatKey or other technology to estimate the standard error from a bootstrap distribution generated from the sample. Then use the standard error to give a \(95 \%\) confidence interval for the proportion of the population to agree with the statement. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. In a random sample of 400 people, 112 agree and 288 disagree.

Construct an interval estimate for the given parameter using the given sample statistic and margin of error. For \(\rho,\) using \(r=0.62\) with margin of error 0.05 .

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