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Chapter 10: Problem 9

Who tweets? Do younger people use Twitter more often than older people? In a random sample of 316 adult Internet users aged 18 to \(29,26 \%\) used Twitter. In a separate random sample of 532 adult Internet users aged 30 to \(49,14 \%\) used Twitter. \({ }\) (a) Calculate the standard error of the sampling distribution of the difference in the sample proportions (younger adults - older adults). What information does this value provide? (b) Construct and interpret a \(90 \%\) confidence interval for the difference between the true proportions of adult Internet users in these age groups who use Twitter.

Short Answer

Expert verified
The standard error is 0.0289. The 90% confidence interval is (0.0725, 0.1675).

Step by step solution

01

Understand the Problem

We have two samples. The first sample has 316 younger adults (aged 18-29) with 26% using Twitter. The second sample has 532 older adults (aged 30-49) with 14% using Twitter. We need to calculate the standard error for the difference in proportions and construct a 90% confidence interval.
02

Calculate Proportions

For the younger group, the sample proportion \( p_1 = 0.26 \). For the older group, the sample proportion \( p_2 = 0.14 \).
03

Formula for Standard Error

The standard error (SE) for the difference in proportions is calculated using the formula: \[SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\]where \(n_1 = 316\) and \(n_2 = 532\).
04

Plug Values into Formula

Substitute the known values into the formula: \[SE = \sqrt{\frac{0.26(1-0.26)}{316} + \frac{0.14(1-0.14)}{532}} \]Calculate each term and sum them up.
05

Calculate SE Value

\[SE = \sqrt{\frac{0.26 \times 0.74}{316} + \frac{0.14 \times 0.86}{532}} \]Calculate each square root term: \[\frac{0.26 \times 0.74}{316} = 0.000608,\frac{0.14 \times 0.86}{532} = 0.000227\]\[SE = \sqrt{0.000608 + 0.000227} = \sqrt{0.000835} = 0.0289\]
06

Information from SE Value

The standard error gives us an estimate of the variation or uncertainty around the difference in the sample proportions. It is used to calculate confidence intervals.
07

Calculate Confidence Interval

To calculate a 90% confidence interval, use the formula:\[(p_1 - p_2) \pm Z^* \cdot SE\]where \(Z^* = 1.645\) for a 90% confidence level.
08

Plug Values into Confidence Interval Formula

\[ (0.26 - 0.14) \pm 1.645 \cdot 0.0289 = 0.12 \pm 0.0475 \]This results in the interval \[0.0725, 0.1675\].
09

Interpret Confidence Interval

We are 90% confident that the true difference in proportions of younger and older internet users who use Twitter is between 7.25% and 16.75%. Younger adults are more likely to use Twitter.

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Key Concepts

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

Standard Error
The standard error (SE) is a crucial concept in statistics that helps us measure the precision of our sample statistic. It quantifies the amount of variability or dispersion we can expect when estimating the population parameter using sample data. In simpler terms, SE allows us to understand how much the sample statistic would fluctuate if we repeatedly took samples from the same population.

When dealing with the difference in proportions between two groups, the SE can be calculated using specific formulas tailored to those conditions. For example, if you're comparing two groups' proportions, the formula looks like:\[SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\]where:
  • \(p_1\) and \(p_2\) are the sample proportions of the two groups, and
  • \(n_1\) and \(n_2\) are the sample sizes of the two groups.
The smaller the SE, the more precise our estimate is, providing a clearer picture of the data relationship between the groups.
Sample Proportion
The sample proportion is a way to express the outcomes of a sample in categorical data as a fraction or percentage. It represents the fraction of the sample that has a certain characteristic. For example, if we want to know how many people in a sample use Twitter, the sample proportion would show us this percentage.

Calculating the sample proportion \(p\) is straightforward:\[p = \frac{x}{n}\]where \(x\) is the number of successful outcomes (e.g., people using Twitter) and \(n\) is the total number of observations in the sample.

This value provides a snapshot of the sample and is fundamental in estimating the population proportion. In our exercise, the younger and older adults' sample proportions help compare the two groups' behaviors, making it possible to understand if one demographic is more inclined to use Twitter than the other.
Difference in Proportions
In statistics, the difference in proportions refers to the subtraction of one sample proportion from another to find what the difference between them is. This value aids in determining if there is a substantial difference between the two groups regarding a specific trait.

For instance, in our context:\[p_1 - p_2 = 0.26 - 0.14 = 0.12\]This result shows that a 12% greater portion of younger adults use Twitter compared to older adults.

The difference in proportions is fundamental when creating confidence intervals. It helps in estimating the range wherein the true difference between population proportions likely resides. This informs decisions or hypotheses about if one group significantly differs from another in a given characteristic.
Random Sampling
Random sampling is a pivotal concept in statistics that ensures every individual in a population has an equal chance of being included in the sample. This practice helps to establish a sample that truly represents the entire population. By using random samples, statisticians can make more accurate generalizations and ensure that bias does not skew the results.

In our exercise, random samples of younger and older adults allow us to confidently compare the Twitter usage between these groups, with the understanding that our samples are good representations of their respective age groups.

Random sampling helps uphold the principles of fairness and unbiased data collection, which are cornerstones of credible statistical analysis. This method is vital for drawing accurate conclusions that can be generalized to the larger population.

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Most popular questions from this chapter

Driving school A driving school owner believes that Instructor \(\mathrm{A}\) is more effective than Instructor \(\mathrm{B}\) at preparing students to pass the state's driver's license exam. An incoming class of 100 students is randomly assigned to two groups, each of size \(50 .\) One group is taught by Instructor \(A ;\) the other is taught by Instructor B. At the end of the course, 30 of Instructor A's students and 22 of Instructor \(\mathrm{B}\) 's students pass the state exam. (a) Do these results give convincing evidence at the \(\alpha=0.05\) level that Instructor \(\mathrm{A}\) is more effective? (b) Describe a Type I and a Type II error in this setting. Which error could you have made in part (a)?

The researchers report that the results were statistically significant at the \(1 \%\) level. Which of the following is the most appropriate conclusion? (a) Because the \(P\) -value is less than \(1 \%\), fail to reject \(H_{0}\). There is not convincing evidence that the proportion of male college students in the study who worked for pay last summer is different from the proportion of female college students in the study who worked for pay last summer. (b) Because the \(P\) -value is less than \(1 \%\), fail to reject \(H_{0}\). There is not convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer.

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 whether they were being prayed for. Explain why this is important.

A better drug? In a pilot study, a company's new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood \((\mathrm{mg} / \mathrm{dl})\) greater than with the current drug, the company will begin the expensive process of mass- producing the new drug. For the 14 subjects who were assigned at random to the current drug. the mean cholesterol reduction was \(54.1 \mathrm{mg} / \mathrm{dl}\) with a standard deviation of \(11.93 \mathrm{mg} / \mathrm{dl}\). For the \(15 \mathrm{sub}\) jects who were randomly assigned to the new drug, the mean cholesterol reduction was \(68.7 \mathrm{mg} / \mathrm{dl}\) with a standard deviation of \(13.3 \mathrm{mg} / \mathrm{dl}\). Graphs of the data reveal no outliers or strong skewness. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not \(\left.H_{0}: \mu_{1}-\mu_{2}=0 .\right)\) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

Exercises 58 to 60 refer to the following setting. A study of road rage asked random samples of \(596 \mathrm{men}\) 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 phone numbers. The researchers performed a test of the following hypotheses: \(H_{0}: \mu_{M}=\mu_{F}\) versus \(H_{a^{*}} \mu_{M} \neq \mu_{F}\) The \(P\) -value for the stated hypotheses is 0.002 . Interpret this value in the context of this study. (a) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means. (b) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference. (c) Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference. (d) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true. (e) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.
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