/*! 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 115 Exercises 3.112 to 3.115 give in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 115

Exercises 3.112 to 3.115 give information about the proportion of a sample that agree with a certain statement. Use StatKey or other technology to find a confidence interval at the given confidence level for the proportion of the population to agree, using percentiles from a bootstrap distribution. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. Find a \(99 \%\) confidence interval if, in a random sample of 1000 people, 382 agree, 578 disagree, and 40 can't decide.

Short Answer

Expert verified
The 99% confidence interval would be the range calculated as [lower limit, upper limit]. To find the precise interval, replace p and SE with their calculated values in the formula to calculate the lower and upper limits.

Step by step solution

01

Calculate the Proportion

Calculate the proportion of people who agreed with the statement by dividing the number of people who agreed by the total number of people who made a decision. So, it will be \(p = \frac{382}{(382 + 578)} = \frac{382}{960}\).
02

Calculate the Standard Error

Next, calculate the standard error using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the total number of responses (excluding those who couldn't decide). So, it would be \(SE = \sqrt{\frac{p(1-p)}{960}}\).
03

Find the Confidence Interval

For a 99% confidence interval, we need to find the 0.5% and 99.5% percentile values from a Normal distribution(Normally these values are -2.576 and 2.576). Then, calculate the confidence limits as \(confidence\_interval = p 卤 Z_{value} * SE\).[lower limit = \(p - 2.576 * SE\) and upper limit = \(p + 2.576 * SE\)]

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 91影视!

Key Concepts

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

Bootstrap Distribution

Understanding the bootstrap distribution is essential for estimating the variability of a statistic, like a sample proportion. The bootstrap distribution is generated by repeatedly resampling from the original data, with replacement, and recalculating the statistic for each resample. This process creates a distribution of the statistic that approximates the sampling distribution we would see if we could take many samples from the population.

In practice, this means that to create a bootstrap distribution for our sample proportion, we repeatedly sample 960 responses (the size of the sample without indecisive respondents) and calculate the proportion that agrees each time. Over thousands of such resamples, we build up a bootstrap distribution that we can use to estimate the variability of our sample proportion.

Proportion

The term 'proportion' refers to the fraction of the total sample that exhibits a particular trait or characteristic. In the problem at hand, we calculated the proportion of people who agree with the statement out of those who made a decision. To find this, we divided the number of people who agreed, 382, by the total number of decisions, which is the sum of agreements and disagreements, 960. This proportion is representative of the larger population and is used to draw inferences about the population's sentiment regarding the statement.

Standard Error

The standard error (SE) measures the amount of variability in the sampling distribution of a statistic. In this context, it gives us an idea of how precisely we have estimated the sample proportion. The standard error is calculated using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the sample proportion and \(n\) is the sample size. A smaller standard error indicates a more precise estimate, which in turn provides a narrower confidence interval, offering a more precise estimate of the population proportion.

Confidence Level

A confidence level indicates the degree of certainty that the true population parameter lies within the calculated confidence interval. For example, a 99% confidence level means that if we were to take many samples and compute a confidence interval for each sample, we would expect about 99% of those intervals to contain the population parameter. Higher confidence levels correspond to wider confidence intervals, reflecting a higher certainty about the estimate including the true population parameter, but they also mean less precision about where that parameter is located.

StatKey Technology

StatKey technology is a set of interactive web-based statistical tools that provides a convenient way for students to perform bootstrap simulations and other statistical analysis. To solve the problem presented, one would use the 'CI for Single Proportion' tool within StatKey. By entering the data into the tool, it calculates a bootstrap distribution for us, from which we can then find the confidence interval for the population proportion at the specified confidence level. It simplifies the process of statistical analysis, making it accessible to students without advanced programming or statistical skills.

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

Downloading Apps for Your Smartphone A random sample of \(n=461\) smartphone users in the US in January 2015 found that 355 of them have downloaded an app. \(^{10}\) (a) Give notation for the parameter of interest, and define the parameter in this context. (b) Give notation for the quantity that gives the best estimate and give its value. (c) What would we have to do to calculate the parameter exactly?

Effect of Overeating for One Month: Correlation between Short-Term and Long- Term Weight Gain In Exercise 3.70 on page \(227,\) we describe a study in which participants ate significantly more and exercised significantly less for a month. Two and a half years later, participants weighed an average of 6.8 pounds more than at the start of the experiment (while the weights of a control group had not changed). Is the amount of weight gained over the following 2.5 years directly related to how much weight was gained during the one-month period? For the 18 participants, the correlation between increase of body weight during the one-month intervention and increase of body weight after 30 months is \(r=0.21 .\) We want to estimate, for the population of all adults, the correlation between weight gain over one month of bingeing and the effect of that month on a person's weight 2.5 years later. (a) What is the population parameter of interest? What is the best estimate for that parameter? (b) To find the sample correlation \(r=0.21,\) we used a dataset containing 18 ordered pairs (weight gain over the one month and weight gain 2.5 years later for each individual in the study). Describe how to use this data to obtain one bootstrap sample. (c) What statistic is recorded for the bootstrap sample? (d) Suppose that we use technology to calculate the relevant statistic for 1000 bootstrap samples. Describe how to find the standard error using those bootstrap statistics. (e) The standard error for one set of bootstrap statistics is 0.14. Calculate a \(95 \%\) confidence interval for the correlation. (f) Use the confidence interval from part (e) to indicate whether you are confident that there is a positive correlation between amount of weight gain during the one-month intervention and amount of weight gained over the next 2.5 years, or whether it is plausible that there is no correlation at all. Explain your reasoning. (g) Will a \(90 \%\) confidence interval most likely be wider or narrower than the \(95 \%\) confidence interval found in part (e)?

\(\mathbf{3 . 1 2 3}\) What Proportion Have Pesticides Detected? In addition to the quantitative variable pesticide concentration, the researchers also report whether or not the pesticide was detected in the urine (at standard detection levels). Before the participants started eating organic, 111 of the 240 measurements (combining all pesticides and people) yielded a positive pesticide detection. While eating organic, only 24 of the 240 measurements resulted in a positive pesticide detection. (a) Calculate the sample difference in proportions: proportion of measurements resulting in pesticide detection while eating non- organic minus proportion of measurements resulting in pesticide detection while eating organic. (b) Figure 3.33 gives a bootstrap distribution for the difference in proportions, based on \(1000 \mathrm{sim}-\) ulated bootstrap samples. Approximate a \(98 \%\) confidence interval. (c) Interpret this interval in context.

Exercises 3.71 to 3.73 consider the question (using fish) of whether uncommitted members of a group make it more democratic. It has been argued that individuals with weak preferences are particularly vulnerable to a vocal opinionated minority. However, recent studies, including computer simulations, observational studies with humans, and experiments with fish, all suggest that adding uncommitted members to a group might make for more democratic decisions by taking control away from an opinionated minority. \({ }^{36}\) In the experiment with fish, golden shiners (small freshwater fish who have a very strong tendency to stick together in schools) were trained to swim toward either yellow or blue marks to receive a treat. Those swimming toward the yellow mark were trained more to develop stronger preferences and became the fish version of individuals with strong opinions. When a minority of five opinionated fish (wanting to aim for the yellow mark) were mixed with a majority of six less opinionated fish (wanting to aim for the blue mark), the group swam toward the minority yellow mark almost all the time. When some untrained fish with no prior preferences were added, however, the majority opinion prevailed most of the time. \({ }^{37}\) Exercises 3.71 to 3.73 elaborate on this study. 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?

Do You Find Solitude Distressing? "For many people, being left alone with their thoughts is a most undesirable activity," says a psychologist involved in a study examining reactions to solitude. \({ }^{26}\) In the study, 146 college students were asked to hand over their cell phones and sit alone, thinking, for about 10 minutes. Afterward, 76 of the participants rated the experience as unpleasant. Use this information to estimate the proportion of all college students who would find it unpleasant to sit alone with their thoughts. (This reaction is not limited to college students: in a follow-up study involving adults ages 18 to 77 , a similar outcome was reported.) (a) Give notation for the quantity being estimated, and define any parameters used. (b) Give notation for the quantity that gives the best estimate, and give its value. (c) Give a \(95 \%\) confidence interval for the quantity being estimated, given that the margin of error for the estimate is \(8 \%\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation
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.