/*! 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 11 Who uses instant messaging? Do y... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 11

Who uses instant messaging? Do younger people use online instant messaging (IM) more often than older people? A random sample of IM users found that 73 of the 158 people in the sample aged 18 to 27 said they used IM more often than email. In the 28 to 39 age group, 26 of 143 people used IM more often than email.\(^{9}\) Construct and interpret a 90% confidence interval for the difference between the proportions of IM users in these age groups who use IM more often than email.

Short Answer

Expert verified
The 90% confidence interval is (0.1807, 0.3793).

Step by step solution

01

Define Parameters

Let \( p_1 \) be the proportion of people aged 18-27 who use instant messaging more often than email, and \( p_2 \) be the proportion of people aged 28-39 who use instant messaging more often than email.
02

Calculate Sample Proportions

Calculate the sample proportion for each age group. The sample proportion for the 18-27 age group is \( \hat{p}_1 = \frac{73}{158} \approx 0.462 \). The sample proportion for the 28-39 age group is \( \hat{p}_2 = \frac{26}{143} \approx 0.182 \).
03

Determine Standard Error

The standard error (SE) for the difference between two proportions is given by:\[ SE = \sqrt{\frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2} } \]Where \( n_1 = 158 \) and \( n_2 = 143 \). Substitute the values to find the SE:\[ SE = \sqrt{\frac{0.462(1-0.462)}{158} + \frac{0.182(1-0.182)}{143} } \approx 0.0604 \]
04

Find the Critical Value

For a 90% confidence interval, the critical value (\( z \)) is approximately 1.645 (using a standard normal distribution table or calculator).
05

Calculate Margin of Error

The margin of error (ME) is given by:\[ ME = z \times SE \]Substitute the values:\[ ME = 1.645 \times 0.0604 \approx 0.0993 \]
06

Construct the Confidence Interval

The confidence interval for the difference in proportions \((p_1 - p_2)\) is given by:\[ (\hat{p}_1 - \hat{p}_2) \pm ME \]Substitute the values:\[ (0.462 - 0.182) \pm 0.0993 = 0.280 \pm 0.0993 \]This gives the interval: (0.1807, 0.3793).
07

Interpret the Confidence Interval

We are 90% confident that the true difference in the proportions of younger (18-27) and older (28-39) IM users who use IM more than email is between 18.07% and 37.93%.

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.

Sampling Distribution
Understanding the concept of the sampling distribution is essential for making inferences about a population using sample data. The sampling distribution refers to the distribution of a statistic (like a sample proportion) calculated from random samples of a certain size drawn from a population. Here, we're dealing with the difference in proportions between two groups.

In this exercise, we take a look at two different age groups of instant messaging users: one group aged 18-27 and the other 28-39. For each group, we've drawn a sample and calculated its proportion of individuals who prefer IM over email. The sampling distribution of the difference in these proportions helps us estimate the true difference in their preferences across the entire population. The shape of this distribution is approximately normal, especially when the sample size is large enough as it is in this study.
Sample Proportion
Let's dive into what the sample proportion means. It's simply the ratio of individuals in a sample with a particular characteristic to the total number of individuals in that sample.

For the 18-27 age group, 73 out of 158 people prefer IM over email, giving us a sample proportion of:
  • \[ \hat{p}_1 = \frac{73}{158} \approx 0.462 \]
For the 28-39 age group, we have 26 out of 143 people, resulting in a sample proportion of:
  • \[ \hat{p}_2 = \frac{26}{143} \approx 0.182 \]
These sample proportions act as our best estimate of the true proportions in the entire population for each group. They are foundational for further calculations like standard error and margin of error.
Standard Error
The standard error measures the variability or spread of the sample statistic鈥攈ere, the difference in proportions鈥攚hen you repeatedly take samples of the same size from the population.

To calculate it for the difference between two sample proportions, use the formula:
  • \[SE = \sqrt{\frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2} }\]
With our data:
  • \( n_1 = 158 \text{ and } n_2 = 143 \)
  • \( SE = \sqrt{\frac{0.462(1-0.462)}{158} + \frac{0.182(1-0.182)}{143} } \approx 0.0604 \)
This value indicates how much we expect the sample difference \( \hat{p}_1 - \hat{p}_2 \) to vary from the true population difference. Smaller standard errors suggest more precise estimates.
Margin of Error
The margin of error provides a range within which we expect the true population parameter (the difference in proportions) to fall, with a certain degree of confidence.

To find this range, multiply the standard error by a critical value from the normal distribution corresponding to the desired confidence level. For a 90% confidence interval, the critical value \( z \) is 1.645.
  • \[ME = z \times SE = 1.645 \times 0.0604 \approx 0.0993\]
So, the margin of error of approximately 0.0993 indicates the span of uncertainty around our sample proportion difference estimate. It expands our confidence interval, making it \((0.1807, 0.3793)\). Hence, we are 90% confident that the true difference in IM usage preference between the younger and older groups is between 18.07% and 37.93%.

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

Information online \((8.2,10.1)\) A random digit dialing sample of 2092 adults found that 1318 used the Internet. \(^{45}\) Of the users, 1041 said that they expect businesses to have Web sites that give product information; 294 of the 774 nonusers said this. (a) Construct and interpret a 95% confidence interval for the proportion of all adults who use the Internet. (b) Construct and interpret a 95% confidence interval to compare the proportions of users and nonusers who expect businesses to have Web sites.

What鈥檚 wrong? 鈥淲ould you marry a person from a lower social class than your own?鈥� Researchers asked this question of a random sample of 385 black, never- married students at two historically black colleges in the South. Of the 149 men in the sample, 91 said 鈥淵es.鈥� Among the 236 women, 117 said 鈥淵es.鈥漒(^{14}\) Is there reason to think that different proportions of men and women in this student population would be willing to marry beneath their class? Holly carried out the significance test shown below to answer this question. Unfortunately, she made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}=\rho_{2}$$ $$H_{a} : p_{1} \neq p_{2}$$ at the 95% confidence level. Plan: If conditions are met, I鈥檒l do a one-sample \(z\) test for comparing two proportions. \(\cdot\) Random The data came from a random sample of 385 black, never-married students. \(\cdot\) Normal One student's answer to the question should have no relationship to another student's answer. \(\cdot\) Independent The counts of successes and falures in the two groups - \(91,58,117,\) and \(119-\) are all at least 10 . Do: From the data, \(\hat{p}_{1}=\frac{91}{149}=0.61\) and \(\hat{p}_{2}=\frac{117}{236}=0.46\) \(\bullet\) Test statistic $$z=\frac{(0.61-0.46)-0}{\sqrt{\frac{0.61(0.39)}{149}+\frac{0.46(0.54)}{236}}}=2.91$$ \(\cdot P\) value From Table \(A, P(z \geq 2.91)=1-0.9982=\) 0.0018 . Conclude: The P-value, \(0.0018,\) is less than \(0.05,\) so I'll reject the null hypothesis. This proves that a higher proportion of men than women are willing to marry someone from a social class lower than their own.

Steroids in high school A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used to improve athletic performance.\(^{13}\) Is there a significant difference between the population proportions? State appropriate hypotheses for a significance test to answer this question. Define any parameters you use.

Paired or unpaired? In each of the following settings, decide whether you should use paired \(t\) procedures or two-sample t procedures to perform inference. Explain your choice.\(^{43}\) (a) To compare the average weight gain of pigs fed two different rations, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Ration A and which got Ration B. (b) A random sample of college professors is taken. We wish to compare the average salaries of male and female teachers. (c) To test the effects of a new fertilizer, 100 plots are treated with the new fertilizer, and 100 plots are treated with another fertilizer. A computer鈥檚 random number generator is used to determine which plots get which fertilizer.

Multiple choice: Select the best answer for Exercises 29 to 32. A sample survey interviews SRSs of 500 female college students and 550 male college students. Each student is asked whether he or she worked for pay last summer. In all, 410 of the women and 484 of the men say 鈥淵es.鈥� Exercises 29 to 31 are based on this survey. The 95\(\%\) confidence interval for the difference \(p_{M}-p_{F}\) in the proportions of college men and women who worked last summer is about (a) \(0.06 \pm 0.00095\) (b) \(0.06 \pm 0.043\) (c) \(0.06 \pm 0.036\) (d) \(-0.06 \pm 0.043\) (e) \(-0.06 \pm 0.036\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Calculus

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.