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Chapter 10: Problem 76

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.

Short Answer

Expert verified
(a) 95% CI for Internet usage: [0.607, 0.645]; (b) CI for expectations difference: [0.135, 0.218].

Step by step solution

01

Identify the sample proportion for all adults using the Internet

We have 1318 adults using the Internet out of a total sample of 2092. The sample proportion \( \hat{p} \) is determined using \( \hat{p} = \frac{1318}{2092} \).
02

Calculate the standard error for the proportion

The standard error (SE) for a proportion is calculated using \( \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is 2092. Substitute the sample proportion from Step 1 into this formula.
03

Determine the 95% confidence interval for Internet usage

The 95% confidence interval is given by \( \hat{p} \pm z_{*} \times \text{SE} \). For a 95% confidence level, \( z_{*} \) is approximately 1.96. Calculate the interval using values from Step 1 and Step 2.
04

Identify sample proportions for users and non-users expecting businesses to have websites

For users, \( \hat{p}_1 = \frac{1041}{1318} \). For non-users, \( \hat{p}_2 = \frac{294}{774} \). These represent the proportions of those expecting websites among users and non-users, respectively.
05

Calculate the standard error for the difference in proportions

The SE for the difference between two independent proportions is given by \( \text{SE}_{difference} = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \), where \( n_1 = 1318 \) and \( n_2 = 774 \). Plug \( \hat{p}_1 \) and \( \hat{p}_2 \) from Step 4 into this formula.
06

Construct the 95% confidence interval for the difference in expectations

Use the formula \( (\hat{p}_1 - \hat{p}_2) \pm z_{*} \times \text{SE}_{difference} \) to determine the interval, where \( z_{*} = 1.96 \). Substitute the calculated proportions and SE from Step 4 and Step 5.

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Key Concepts

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

Proportion Calculations
Proportion calculations are a key part of understanding survey data. When you want to know how many people use the Internet in a group, you calculate the proportion. The formula is simple: find the number of people who meet your criteria and divide that by the total number of people you surveyed.
In the exercise provided, 1318 out of 2092 adults use the Internet. By placing these numbers into our formula, the sample proportion is determined: \( \hat{p} = \frac{1318}{2092} \). This results in a proportion of about 0.63 or 63%.
This tells us that 63% of the adults in our sample use the Internet. Proportion calculations like these are foundational for further statistical analysis.
Standard Error
The standard error (SE) is all about understanding the accuracy of your sample proportion. Essentially, it reveals how much your sample result might vary from the actual population result.
The formula for the standard error when dealing with proportions looks like this: \( \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion and \( n \) is the sample size.
Using our proportion from earlier (0.63) and the total number of sampled people (2092), the SE helps us determine if the sample proportion accurately reflects the broader population. A smaller SE indicates a more precise estimate, which is crucial for creating reliable confidence intervals.
Internet Usage Statistics
Internet usage statistics reveal insights into how frequently and broadly people access the Internet. By calculating these statistics, like in the provided exercise, we can tell how many adults in the survey rely on the Internet.
With 63% of adults using the Internet according to our calculated proportion, these stats can guide businesses in deciding where to allocate resources, such as enhancing online presence.
Moreover, when we examine expectations for businesses to have websites, we draw further inferences. These statistics show differences in attitudes between Internet users and non-users, highlighting the digital divide that may exist in perceptions and expectations.
Statistical Inference in Psychology
Statistical inference plays a huge role in psychology, helping researchers draw conclusions about behaviors and attitudes in populations. For this to be effective, understanding the margin of error and confidence intervals is essential.
The psychological study of Internet users can reveal patterns in behavior or expectations, as evident from our exercise. When comparing Internet users and non-users' expectations from businesses, we see different psychological profiles emerging.
Through calculating the difference in expectations and constructing confidence intervals, researchers can infer psychological trends and create more targeted interventions or studies. This approach ensures findings in psychology are grounded in robust statistical evidence, adding reliability and validity.

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Most popular questions from this chapter

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) \(0.001 < P < 0.002\)

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 . Broken crackers We don鈥檛 like to find broken crackers when we open the package. How can makers reduce breaking? One idea is to microwave the crackers for 30 seconds right after baking them. Breaks start as hairline cracks called 鈥渃hecking.鈥� Assign 65 newly baked crackers to the microwave and another 65 to a control group that is not microwaved. After one day, none of the microwave group and 16 of the control group show checking.\(^{8}\)

Down the toilet A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not \(H_{0} : \mu_{1}-\mu_{2}=0\) ) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 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鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he 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 $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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 pooled sample proportion who worked last summer is about (a) \(\hat{p}_{\mathrm{C}}=1.70 . \quad(\mathrm{d}) \hat{p}_{\mathrm{C}}=0.85\) (b) \(\hat{p}_{\mathrm{C}}=0.89 . \quad\) (e) \(\hat{p}_{\mathrm{C}}=0.82\) (c) \(\hat{p}_{\mathrm{C}}=0.88\)
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