/*! 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 19 Refer to Exercise 18 (Section 8.... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 19

Refer to Exercise 18 (Section 8.5). The percentage of people catching a cold when exposed to a cold virus is shown in the table, for a group of people with only a few social contacts and a group with six or more activities. \(^{21}\) $$\begin{array}{lcc}\hline & \text { Few Social Outlets } & \text { Many Social Outlets } \\\\\hline \text { Sample Size } & 96 & 105 \\\\\text { Percent with Colds } & 62 \% & 35 \% \\\\\hline\end{array}$$ Construct a \(95 \%\) lower confidence bound for the difference in population proportions. Does it appear that when exposed to a cold virus, a greater proportion of those with fewer social contacts contracted a cold? Is this counter-intuitive?

Short Answer

Expert verified
Answer: Yes, with a 95% confidence interval, it appears that those with fewer social outlets have a higher proportion of colds when exposed to the cold virus, as the lower bound of the difference in population proportions is approximately 0.1334 (positive).

Step by step solution

01

Organize the data

Organize the given data into a clear table format, which is helpful to work with. | Group | Sample Size (n) | Percent with Colds (p) | |-------|-----------------|------------------------| | Few Social Outlets (1) | 96 | 0.62 | | Many Social Outlets (2) | 105 | 0.35 |
02

Calculate sample proportions

Calculate the sample proportions for each group using Sample Size and Percent with Colds data. For Few Social Outlets: \(p_1 = 0.62\) For Many Social Outlets: \(p_2 = 0.35\)
03

Calculate the difference in sample proportions

Find the difference in sample proportions by subtracting \(p_2\) from \(p_1\). \(\hat{p} = p_1 - p_2 = 0.62 - 0.35 = 0.27\)
04

Compute the standard error of the difference in proportions

Calculate the standard error of the difference. The formula for the standard error is: \(SE(\hat{p}) = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\) \(SE(\hat{p}) = \sqrt{\frac{0.62(1-0.62)}{96} + \frac{0.35(1-0.35)}{105}}\) \(SE(\hat{p}) \approx 0.0805\)
05

Calculate the 95% lower confidence bound

We'll use the 95% confidence level, which corresponds to a z-score of 1.645 for a one-tailed test (since we're interested in the lower bound). The formula for the confidence interval is: \(CI = \hat{p} - z * SE(\hat{p})\) \(CI = 0.27 - 1.645 * 0.0805 \approx 0.1334\)
06

Interpret the results

The 95% lower confidence bound for the difference in population proportions is approximately 0.1334. Since the confidence interval is positive, it implies that a greater proportion of those with fewer social contacts contracted a cold when exposed to the cold virus. This result may be counter-intuitive, as we might expect that people with fewer social contacts would have a lower risk of catching a cold.

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Key Concepts

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

Confidence Interval
A confidence interval gives us a range of values that is likely to contain the true difference in population proportions. This interval is built around our estimated difference in sample proportions and is extended on either side by a margin of error. The confidence level, often expressed as a percentage, indicates how confident we are that the true population parameter is within this range.

In this scenario, we have a 95% confidence interval, meaning we are 95% sure that the true difference in population proportions of catching a cold between two groups falls within this interval. We calculated this interval by determining the point estimate of the difference and adjusting for variability in the sample data.
  • Confidence intervals help us understand the reliability of our estimates.
  • In this case, the interval suggests that the true difference is indeed positive.
This provides insight into the potential underlying differences in susceptibility in both populations.
Sample Proportion
The sample proportion refers to the fraction of individuals in a sample that exhibit a specific characteristic鈥攊n this case, catching a cold. We calculate it by dividing the number of individuals with the characteristic by the total sample size for each group.

For instance, the sample proportion of those with colds in the 'Few Social Outlets' group is given by \[ p_1 = \frac{59.52}{96} = 0.62 \]This means 62% of the people in this sample got colds. Similarly, for the 'Many Social Outlets' group, \[ p_2 = \frac{36.75}{105} = 0.35 \]This calculation shows that 35% of the people in this group got colds.
  • The difference in these proportions is crucial for understanding contrasts between the two groups.
  • It lays the groundwork for constructing the confidence interval.
Sample proportions thus serve as the starting point for assessing the possible population differences.
Standard Error
The standard error of the difference in proportions measures the variability or spread of our sample distribution. It plays a crucial role in the creation of confidence intervals and hypothesis testing.

The formula used is:\[ SE(\hat{p}) = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \]This formula accounts for the variability in each sample group. By calculating the standard error, we can adjust our estimated difference to better reflect the sample data's variability.
  • The smaller the standard error, the more precise our difference estimate.
  • It signifies how much the sample proportions vary.
A standard error, such as 0.0805 in our example, allows us to construct a confidence interval by using it to calculate the margin of error.
Population Proportion Difference
The population proportion difference is a measure of how two groups differ in the proportion with a certain characteristic鈥攊n this case, the tendency to catch a cold. The difference in sample proportions is an estimate of this true, but unknown, difference in populations.

In our example, we calculated the difference in sample proportions as:\[ \hat{p} = p_1 - p_2 = 0.62 - 0.35 = 0.27 \]This difference suggests that there is typically a 27% higher incidence of colds in the 'Few Social Outlets' group compared to the 'Many Social Outlets' group when exposed to a virus.
  • Understanding this difference helps us address crucial questions about behavior and health.
  • This can be counter-intuitive, suggesting exposure and social activity dynamics are complex.
Knowing this difference helps guide further investigation and provides a basis for understanding why such differences might occur.

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Most popular questions from this chapter

Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

Calculate the margin of error in estimating a population mean \(\mu\) for the values given in Exercises 3-6. Comment on how a larger population variance affects the margin of error: \(n=30, \sigma^{2}=1.5\)

Catching a Cold Do well-rounded people get fewer colds? A study in the Chronicle of Higher Education found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities. \({ }^{21}\) Suppose that of the 276 healthy men and women tested, \(n_{1}=96\) had only a few social outlets and \(n_{2}=105\) were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: \begin{tabular}{lcc} \hline & Few Social Outlets & Many Social Outlets \\ \hline Sample Size & 96 & 105 \\ Percent with Colds & \(62 \%\) & \(35 \%\) \\ \hline \end{tabular} a. Construct a \(99 \%\) confidence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected finding?

Use the information given to find the necessary confidence interval for the binomial proportion \(p .\) Interpret the interval that you have constructed. A \(90 \%\) confidence interval for \(p\), based on a random sample of \(n=300\) observations from a binomial population with \(x=263\) successes.

Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\)
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