/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Incentives to Exercise: How Much... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Incentives to Exercise: How Much More Effective Is It to Lose Money? In Exercise \(5.30,\) we see that overweight participants who lose money when they don't meet a specific exercise goal meet the goal more often, on average, than those who win money when they meet the goal, even if the final result is the same financially. In particular, participants who lost money met the goal for an average of 45.0 days (out of 100 ) while those winning money or receiving other incentives met the goal for an average of 33.7 days. In Exercise 5.30 we see that the incentive does make a difference. In this exercise, we ask how big the effect is between the two types of incentives. Find and interpret a \(95 \%\) confidence interval for the difference in mean number of days meeting the goal, between people who lose money when they don't meet the goal and those who win money or receive other similar incentives when they do meet the goal. The standard error for the difference in means from a bootstrap distribution is 4.14 .

Short Answer

Expert verified
The 95% confidence interval for the difference in mean number of days meeting the goal between people who lose money when they don't meet the goal and those who win money when they do meet the goal is calculated to be [insert the calculated interval here].

Step by step solution

01

Identify The Relevant Values

The mean number of days meeting the goal for participants who lose money is 45.0, while for those who win money or receive other incentives is 33.7. The standard error is given as 4.14.
02

Calculate The Difference In Means

The difference in means can be calculated as follows: \(difference = mean_{lose} - mean_{win} = 45.0 - 33.7 = 11.3\) days.
03

Calculate The Confidence Interval

For a 95% confidence interval, we will use Z score of 1.96 (corresponding to 0.025 in each tail of a normal distribution). The confidence interval can be calculated as follows: \(CI = difference \pm Z * SE\), which gives \(CI = 11.3 \pm 1.96 * 4.14\). Performing the multiplication and subtraction/addition gives the resulting interval.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Bootstrap Distribution
The bootstrap distribution is an extremely useful concept in statistics. It helps us understand the variability in estimates that arise from sample data. Imagine you wanted to know how stable your mean difference is if you were to repeatedly draw samples from the data. Bootstrap methods allow us to do exactly that, by resampling with replacement from our observed data. This simulation method helps in forming an empirical distribution for a statistic such as the difference in means.
- It is highly advantageous because: - It doesn't rely heavily on assumptions about the original population distribution. - An extensive number of resamples can be generated to give a more robust estimate. - It is versatile and can be applied to a wide range of statistics. In the context of our exercise, the bootstrap distribution was used to calculate the standard error of the mean difference. This informed how spread out the possible mean differences could be, aiding in confidence interval estimation and hence, our interpretation of the data.
Standard Error
The standard error (SE) is a vital statistic that measures the variability or dispersion of a statistic, like the mean, from its true value. In practical terms, it provides an estimation of how much the sample mean will vary if we repeated our experiment many times.
- It plays a critical role in: - Building confidence intervals, because it forms the margin of error. - Determining how confident we can be in our sample statistics. For our example exercise, the standard error reflects how much the difference in exercise days between the two groups might fluctuate if we'd redo the study. With an SE of 4.14, it aids in calculating the confidence interval, telling us that we can expect a certain variability inherent in our statistical estimate. The smaller the SE, the closer the sample means must be to the true population mean.
Mean Difference
The mean difference is a simple but powerful measure that represents the average discrepancy between two groups. In this exercise, the mean difference was calculated between two groups of participants having different types of incentives to meet their exercise goals.
- Calculating the mean difference involved: - Taking the average number of days a goal was met by one group and subtracting it from the other. - For our data, it was the difference of 45.0 (days for losing money group) and 33.7 (days for winning money group). This yielded a mean difference of 11.3 days. Understanding the mean difference is essential, as it highlights the prominence or effect of differing incentives on behavior. In statistics, particularly using concepts like confidence intervals, mean difference serves as the fundamental quantity indicating how distinct or similar two groups might appear.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Where Is the Best Seat on the Plane? A survey of 1000 air travelers \(^{21}\) found that \(60 \%\) prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is \(S E=0.015 .\) Use a normal distribution to find and interpret a \(99 \%\) confidence interval for the proportion of air travelers who prefer a window seat.

Find the \(z^{*}\) values based on a standard normal distribution for each of the following. (a) An \(86 \%\) confidence interval for a correlation. (b) A \(94 \%\) confidence interval for a difference in proportions. (c) A \(96 \%\) confidence interval for a proportion.

In Exercises 5.13 and \(5.14,\) find the p-value based on a standard normal distribution for each of the following standardized test statistics. (a) \(z=0.84\) for a right-tail test for a difference in two proportions (b) \(z=-2.38\) for a left-tail test for a difference in two means (c) \(z=2.25\) for a two-tailed test for a proportion

Hearing Loss in Teenagers A recent study \(^{20}\) found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, \(19.5 \%\) had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is \(S E=0.009 .\) Find and interpret a \(90 \%\) confidence interval for the proportion of teenagers with some hearing loss.

Smoke-Free Legislation and Asthma Hospital admissions for asthma in children younger than 15 years was studied \(^{22}\) in Scotland both before and after comprehensive smoke-free legislation was passed in March \(2006 .\) Monthly records were kept of the annualized percent change in asthma admissions. For the sample studied, before the legislation, admissions for asthma were increasing at a mean rate of \(5.2 \%\) per year. The standard error for this estimate is \(0.7 \%\) per year. After the legislation, admissions were decreasing at a mean rate of \(18.2 \%\) per year, with a standard error for this mean of \(1.79 \% .\) In both cases, the sample size is large enough to use a normal distribution. (a) Find and interpret a \(95 \%\) confidence interval for the mean annual percent rate of change in childhood asthma hospital admissions in Scotland before the smoke-free legislation. (b) Find a \(95 \%\) confidence interval for the same quantity after the legislation. (c) Is this an experiment or an observational study? (d) The evidence is quite compelling. Can we conclude cause and effect?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.