/*! 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 97 Give information about the propo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(95 \%\) confidence interval if 180 agree in a random sample of 250 people.

Short Answer

Expert verified
After following these steps and using a software like StatKey, we can find the required confidence interval for the population proportion that agrees with the statement. The exact calculation will depend on the software procedure, but conceptually this is how it can be done.

Step by step solution

01

Calculate Sample Proportion

From the 250 people that were sampled, 180 agreed with the statement. Therefore, the sample proportion is 180/250 = 0.72
02

Use Technology to Find Confidence Interval

Based on our basic statistical knowledge, we need to find confidence interval for population proportion. But in this case we can use technology (StatKey or others) which is much faster and easier. In StatKey, use 'One Proportion', enter the above values : success (people agreed) as 180, total as 250 and set Confidence level to 95%.
03

Analyze the Result

The software will perform a bootstrap method to simulate a sampling distribution and will provide a confidence interval for the population proportion, given as two values: a lower limit and an upper limit.

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 Method
The bootstrap method is a powerful statistical tool used to estimate the distribution of a statistic by resampling with replacement from the original data. This approach allows us to approximate the uncertainty of our estimate without making strict assumptions about the distribution of the population.

When it comes to the confidence interval for a population proportion, the bootstrap method involves repeatedly, say thousands of times, taking samples of the same size as the original from the sampled data. For each bootstrap sample, we calculate the sample proportion that agrees with a certain statement. With these proportions, we build a bootstrap distribution and then determine the confidence interval by finding the appropriate percentiles.

For example, if we desire a 95% confidence interval, we find the percentiles that cut off the lowest 2.5% and highest 2.5% of the bootstrap distribution. The values at these percentiles are the endpoints of the confidence interval. This does not rely on the data following a normal distribution, making the bootstrap method remarkably versatile and robust for calculating confidence intervals.
Sample Proportion
The sample proportion is a key aspect of inferential statistics and is used to estimate the proportion of a population that possesses a certain attribute, based on the data from a sample. It is calculated by dividing the number of observations that have the attribute of interest by the total number of observations in the sample.

In the context of our exercise, the sample proportion (\( p \) ) is the fraction of people in the sample who agree with a particular statement. For example, with 180 people agreeing out of a sample of 250, the sample proportion is \( \frac{180}{250} = 0.72 \), suggesting that 72% of the sampled individuals agree with the statement. This proportion, while a valuable estimate, is subject to sampling variability, hence the importance of computing a confidence interval to better understand the potential range of the true population proportion.
StatKey Technology
StatKey is a suite of web-based statistical tools designed to make modern methods such as bootstrapping and randomization tests accessible to introductory statistics students. It provides a user-friendly interface, which requires no complex programming skills, for performing a variety of statistical analyses.

When calculating confidence intervals, especially for a single proportion, StatKey technology simplifies the process remarkably. Users can easily enter their sample data, specify the type of confidence interval, and immediately receive a visual representation of the bootstrap distribution and the calculated interval. There's no need to perform the complex calculations manually; StatKey automates them and is particularly useful for making the concept of confidence intervals tangible and comprehensible for learners.
Statistical Significance
Statistical significance plays a crucial role in hypothesis testing and helps us determine whether a result is likely due to chance or represents true effects in the population. A result is considered statistically significant if its associated probability of occurring by random chance is lower than a pre-determined significance level, typically 5% (\( p < 0.05 \) ).

With regards to confidence intervals, statistical significance is intertwined with the interval's range. If a 95% confidence interval does not contain the null value (often 0 for differences, 1 for ratios), we assume statistical significance for the test at the 5% level. Thus, the confidence interval gives not just a range of plausible values for our population parameter but also informs us about statistical significance without performing a separate hypothesis test.

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

Give the correct notation for the quantity described and give its value. Correlation between points and penalty minutes for the 24 players with at least one point on $$ \begin{aligned} &\text { Table 3.4 Points and penalty minutes for the 2009-2010 Ottawa Senators NHL team }\\\ &\begin{array}{lllllllllll} \hline \text { Points } & 71 & 57 & 53 & 49 & 48 & 34 & 32 & 29 & 28 & 26 & 26 & 26 \\ \text { Pen mins } & 22 & 20 & 59 & 54 & 34 & 18 & 38 & 20 & 28 & 121 & 53 & 24 \\ \hline \text { Points } & 24 & 22 & 18 & 16 & 14 & 14 & 13 & 13 & 11 & 5 & 3 & 3 \\ \text { Pen mins } & 45 & 175 & 16 & 20 & 20 & 38 & 107 & 22 & 190 & 40 & 12 & 14 \\ \hline \end{array} \end{aligned} $$ the \(2009-2010\) Ottawa Senators \(^{9}\) NHL hockey team. The data are given in Table 3.4 and the full data are available in the file OttawaSenators.

Topical Painkiller Ointment The use of topical painkiller ointment or gel rather than pills for pain relief was approved just within the last few years in the US for prescription use only. \(^{12}\) Insurance records show that the average copayment for a month's supply of topical painkiller ointment for regular users is \(\$ 30 .\) A sample of \(n=75\) regular users found a sample mean copayment of \(\$ 27.90\). (a) Identify each of 30 and 27.90 as a parameter or a statistic and give the appropriate notation for each. (b) If we take 1000 samples of size \(n=75\) from the population of all copayments for a month's supply of topical painkiller ointment for regularusers and plot the sample means on a dotplot, describe the shape you would expect to see in the plot and where it would be centered. (c) How many dots will be on the dotplot you described in part (b)? What will each dot represent?

Student Misinterpretations Suppose that a student is working on a statistics project using data on pulse rates collected from a random sample of 100 students from her college. She finds a \(95 \%\) confidence interval for mean pulse rate to be \((65.5,\) 71.8). Discuss how each of the statements below would indicate an improper interpretation of this interval. (a) I am \(95 \%\) sure that all students will have pulse rates between 65.5 and 71.8 beats per minute. (b) I am \(95 \%\) sure that the mean pulse rate for this sample of students will fall between 65.5 and 71.8 beats per minute. (c) I am \(95 \%\) sure that the confidence interval for the average pulse rate of all students at this college goes from 65.5 to 71.8 beats per minute. (d) I am sure that \(95 \%\) of all students at this college will have pulse rates between 65.5 and 71.8 beats per minute. (e) I am \(95 \%\) sure that the mean pulse rate for all US college students is between 65.5 and 71.8 beats per minute. (f) Of the mean pulse rates for students at this college, \(95 \%\) will fall between 65.5 and 71.8 beats per minute. (g) Of random samples of this size taken from students at this college, \(95 \%\) will have mean pulse rates between 65.5 and 71.8 beats per minute.

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?

What Proportion of Adults and Teens Text Message? A study of \(n=2252\) adults age 18 or older found that \(72 \%\) of the cell phone users send and receive text messages. \(^{14}\) A study of \(n=800\) teens age 12 to 17 found that \(87 \%\) of the teen cell phone users send and receive text messages. What is the best estimate for the difference in the proportion of cell phone users who use text messages, between adults (defined as 18 and over) and teens? Give notation (as a difference with a minus sign) for the quantity we are trying to estimate, notation for the quantity that gives the best estimate, and the value of the best estimate. Be sure to clearly define any parameters in the context of this situation.

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.