/*! 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 15 Who owns iPods? As part of the P... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 15

Who owns iPods? As part of the Pew Internet and American Life Project, researchers surveyed a random sample of 800 teens and a separate random sample of 400 young adults. For the teens, 79% said that they own an iPod or MP3 player. For the young adults, this figure was 67%. 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.

Short Answer

Expert verified
There is a significant difference between the proportions of teens and young adults who own iPods or MP3 players.

Step by step solution

01

Define Parameters

Let \( p_1 \) represent the proportion of teens who own an iPod or MP3 player and \( p_2 \) represent the proportion of young adults who own an iPod or MP3 player.
02

State Hypotheses

We are testing whether there is a significant difference between \( p_1 \) and \( p_2 \). The null hypothesis \( H_0 \) is that there is no difference in the population proportions, i.e., \( p_1 = p_2 \). The alternative hypothesis \( H_a \) is that there is a difference, i.e., \( p_1 eq p_2 \).
03

Calculate Sample Proportions

The sample proportion for teens, \( \hat{p_1} = \frac{79}{100} = 0.79 \). The sample proportion for young adults, \( \hat{p_2} = \frac{67}{100} = 0.67 \).
04

Determine Pooled Proportion

The pooled proportion \( \hat{p} \) is calculated as follows: \( \hat{p} = \frac{(0.79 \times 800) + (0.67 \times 400)}{800 + 400} = 0.75 \).
05

Compute Standard Error

The standard error (SE) of the difference in proportions is given by \( SE = \sqrt{ \hat{p}(1-\hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } \). Substituting the values, \( SE = \sqrt{0.75 \times 0.25 \times \left( \frac{1}{800} + \frac{1}{400} \right)} = 0.0272 \).
06

Calculate Test Statistic

The test statistic \( z \) for comparing the two proportions is given by \( z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.79 - 0.67}{0.0272} = 4.41 \).
07

Make a Conclusion

Since the calculated \( z \) value of 4.41 is greater than the critical value of 1.96 for a 95% confidence level, we reject the null hypothesis. Thus, there is a significant difference between the proportion of teens and young adults who own iPods or MP3 players.

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.

Population Proportions
Population proportions are the fractions of a population exhibiting a certain characteristic. In this exercise, we're interested in the proportion of teens and young adults who own an iPod or MP3 player.
For teens, the population proportion is represented by \( p_1 \), while for young adults, it鈥檚 \( p_2 \).
Understanding these proportions helps in determining how common it is for individuals in these age groups to own such devices.
The key is to compare these proportions from the samples to see if they reflect real differences in the broader populations. This comparison is useful in surveys and market research to understand trends and preferences.
Significance Test
A significance test is a statistical method used to determine whether the observed differences between groups (in this case, population proportions) are genuine or could have occurred by random chance.
In our scenario, we are testing the claim about the difference between teen and young adult ownership of iPods or MP3 players.
  • The null hypothesis \( H_0 \) posits that no difference exists: \( p_1 = p_2 \).
  • The alternative hypothesis \( H_a \) suggests a difference does occur: \( p_1 eq p_2 \).
This process enables us to make informed conclusions about the wider population using sample data.
Standard Error
Standard error (SE) quantifies the amount of variation in a sampling distribution. It's crucial for evaluating how sample data compares to the true population.
Specifically, when comparing two proportions, the standard error helps us understand the extent to which the observed difference might deviate from the true population difference by chance.
Calculation of SE uses the pooled proportion \( \hat{p} \) and both sample sizes:
  • The formula is: \( SE = \sqrt{ \hat{p}(1-\hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } \).
  • This equation considers both sample groups and their respective sizes.
This measure ensures that our test statistic accurately reflects any real differences in the sample data.
Test Statistic
The test statistic is a calculated value used to determine the significance of the test. It converts the observed difference between sample proportions into a measure that can be compared to a known distribution.
In this case, the test statistic is Z, calculated using:
  • The formula: \( z = \frac{\hat{p_1} - \hat{p_2}}{SE} \).
  • This determines how many standard errors the observed difference \( \hat{p_1} - \hat{p_2} \) is away from zero.
A higher absolute value of \( z \) indicates a lesser likelihood that the observed difference is due to random chance alone.
In our example, a \( z \)-value greater than the critical value leads us to reject the null hypothesis, concluding there's a significant difference between the groups.

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

Explain why the conditions for using two-sample z procedures to perform inference about \(p_{1}-p_{2}\) are not met in the settings of Exercises 7 through 10 . Don鈥檛 drink the water! The movie A Civil Action (Touchstone Pictures, 1998) tells the story of a major legal battle that took place in the small town of Woburn, Massachusetts. A town well that supplied water to eastern Woburn residents was contaminated by industrial chemicals. During the period that residents drank water from this well, 16 of the 414 babies born had birth defects. On the west side of Woburn, 3 of the 228 babies born during the same time period had birth defects.

Multiple choice: Select the best answer for Exercises 67 to 70. Exercises 69 and 70 refer to the following setting. A study of road rage asked samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of telephone numbers. The two-sample t statistic for the road rage study (male mean minus female mean) is \(t=3.18\). The \(P\)-value for testing the hypotheses from the previous exercise satisfies (a) \(0.001 < P < 0.005 . \quad\) (d) \(0.002 < P < 0.01\) (b) \(0.0005 < P < 0.001 . \quad(\mathrm{e}) P > 0.01\) (c) \(0.001 < P < 0.002\)

Exercises 23 through 26 involve the following setting. Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves injecting a fertilized egg into the woman鈥檚 uterus. Prayer and pregnancy Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Women in the treatment group were intentionally prayed for by several people (called intercessors) who did not know them, a process known as intercessory prayer. The praying continued for three weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group.\(^{17}\) Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of \(H_{0} : p_{1}=p_{2}\) versus \(H_{a} : p_{1}>p_{2},\) where \(p_{1}\) and \(p_{2}\) are the actual pregnancy rates for women like those in the study who do and don't receive intercessory prayer, respectively. (a) Name the appropriate test and check that the conditions for carrying out this test are met. (b) The appropriate test from part (a) yields a P-value of 0.0007. Interpret this P-value in context. (c) What conclusion should researchers draw at the \(\alpha=0.05\) significance level? Explain. (d) The women in the study did not know if they were being prayed for. Explain why this is important.

Is red wine better than white wine? Observational studies suggest that moderate use of alcohol by adults reduces heart attacks and that red wine may have special benefits. One reason may be that red wine contains polyphenols, substances that do good things to cholesterol in the blood and so may reduce the risk of heart attacks. In an experiment, healthy men were assigned at random to drink half a bottle of either red or white wine each day for two weeks. The level of polyphenols in their blood was measured before and after the two-week period. Here are the percent changes in level for the subjects in both groups:\(^{31}\) Red wine: 3.58 .1\(\quad 7.44 .0 \quad 0.74 .98 .4 \quad 7.05 .5\) White wine: \(3.1 \quad 0.5-3.8\) 4.1 \(-0.62 .7 \quad 1.9-5.9 \quad 0.1\) (a) A Fathom dotplot of the data is shown below. Use the graph to answer these questions: \(\bullet\) Are the centers of the two groups similar or different? Explain. \(\bullet\) Are the spreads of the two groups similar or different? Explain. (b) Construct and interpret a 90% confidence interval for the difference in mean percent change in polyphenol levels for the red wine and white wine treatments. (c) Does the interval in part (b) suggest that red wine is more effective than white wine? Explain.

Did the random assignment work? A large clinical trial of the effect of diet on breast cancer assigned women at random to either a normal diet or a low-fat diet. To check that the random assignment did produce comparable groups, we can compare the two groups at the start of the study. Ask if there is a family history of breast cancer: 3396 of the \(19,541\) women in the low-fat group and 4929 of the \(29,294\) women in the control group said "Yes."\(^{15}\) If the random assignment worked well, there should not be a significant difference in the proportions with a family history of breast cancer. (a) How significant is the observed difference? Carry out an appropriate test to help answer this question. (b) Describe a Type I and a Type II error in this setting. Which is more serious? Explain.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

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.