Question

Do well-rounded people get fewer colds? A study on the Chronicle of Higher \(E d u\) cation was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities. \(^{14}\) 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: $$\begin{array}{lcc} & \text { Few Social Outlets } & \text { Many Social Outlets } \\\\\hline \text { Sample Size } & 96 & 105 \\\\\text { Percent with Colds } & 62 \% & 35 \%\end{array}$$ 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?

Short answer

In addition, provide a possible explanation for the unexpected finding that people with more social outlets have fewer colds. Answer: Yes, there appears to be a significant difference in the population proportions for individuals with few social outlets compared to those with many social outlets, since the 99% confidence interval for the difference in population proportions (0.124, 0.416) does not contain zero. One possible explanation for the unexpected finding is that people with more social outlets may have stronger immune systems because of increased exposure to various pathogens or lower stress levels, which in turn reduces the risk of developing illnesses like the common cold.

Step-by-step solution

Calculate the sample proportions and their difference

First, find the sample proportions \(p_1\) and \(p_2\). For few social outlets, 62% of 96 people got colds; for many social outlets, 35% of 105 people got colds. Calculate the proportions: \(p_1= \frac{62\% \times 96}{100}=\frac{59.52}{96}=0.62\) \(p_2=\frac{35\% \times 105}{100}=\frac{36.75}{105}=0.35\) Then, calculate the difference of proportions: \(\hat{p}_d = p_1 - p_2 = 0.62 - 0.35 = 0.27\)

Calculate the standard error

Next, calculate the standard error of the difference between the sample proportions. The standard error can be determined using the formula: \(SE(\hat{p}_d)= \sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}\) Substitute the values of \(p_1\), \(n_1\), \(p_2\), and \(n_2\): \(SE(\hat{p}_d)= \sqrt{\frac{(0.62)(1-0.62)}{96}+\frac{(0.35)(1-0.35)}{105}} = \sqrt{0.0036} \approx 0.060\)

Find the critical value for a 99% confidence interval

Find the critical value for a 99% confidence interval using a Z-table or calculator. For a two-tailed test, you want to find the Z-score corresponding to an area of \((1-0.99)/2=0.005\) in each tail. The critical value is \(Z_{\alpha/2} = 2.576\)

Calculate the confidence interval

Now, we can use the critical value, the calculated difference, and the standard error to find the confidence interval. Lower limit: \(\hat{p}_d - Z_{\alpha/2} \times SE(\hat{p}_d) = 0.27 - (2.576)(0.06) \approx 0.124\) Upper limit: \(\hat{p}_d + Z_{\alpha/2} \times SE(\hat{p}_d) = 0.27 + (2.576)(0.06) \approx 0.416\) Thus, the 99% confidence interval for the difference in population proportions is \((0.124, 0.416)\). #b. Does there appear to be a difference in the population proportions for the two groups?#

Check if the confidence interval contains zero

Since the 99% confidence interval for the difference in population proportions \((0.124, 0.416)\) does not contain zero, we can conclude that there is a significant 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?# One possible explanation for this unexpected finding is that people with more social outlets may have stronger immune systems due to increased exposure to various pathogens. This exposure may help their immune systems build up a stronger defense against viruses like the common cold. Another possibility is that individuals with more social outlets may have lower levels of stress, which can lead to a stronger immune system and a decreased risk of developing illness.

Key concepts

Difference in Proportions

Understanding the difference in proportions is crucial when comparing two distinct groups. In this context, it involves comparing the likelihood of an event happening in two different sample groups.
For example, if you want to see if people with fewer social activities catch colds more often than those with many, you'd calculate the proportion for each group and then find the difference.

This difference shows whether one group has a significantly higher or lower proportion compared to the other.

• Calculate each sample proportion. For instance, if 62% of people with few social outlets got colds, this proportion is 0.62, while 35% of socially active people results in a proportion of 0.35.
• The difference in proportions would be 0.62 - 0.35 = 0.27.

This difference helps in forming a basis for statistical tests and confidence intervals, providing insight into the likelihood of observing such a difference purely by chance.

Standard Error

The standard error (SE) is a measure of the variability or precision of a sample statistic, such as the difference in proportions in our study.
It's an essential part of constructing confidence intervals and conducting hypothesis tests.

To find the SE for the difference in proportions, use this formula:

• \[ SE(\hat{p}_d) = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \]

Where \(p_1\) and \(p_2\) are the sample proportions, and \(n_1\) and \(n_2\) are the sample sizes.

For instance, using \(0.62\) and \(0.35\) with sample sizes 96 and 105, respectively, the SE can be calculated to understand how much variation in the difference can be expected due to sampling error.
This plays a vital role in determining how much we can trust the difference observed between the two sample groups.

Z-score

The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values.
In confidence interval calculations, the Z-score represents the number of standard deviations a data point is from the mean.

For a 99% confidence interval, the critical Z-score is commonly used to estimate the range within which the true difference in proportions is likely to fall.

• This value is typically found using a Z-table, representing the tails of a standard normal distribution.
• For a 99% confidence level, the critical Z-score is often 2.576.

By applying this critical value to the standard error and difference in proportions, you can calculate the boundaries of the confidence interval, assessing how likely a measured difference represents the true population difference.

Population Proportion

Population proportion refers to the fraction of the total population that possesses a certain attribute or characteristic.
This fulfills a foundational role in statistics, especially when making inferences about larger groups based on sample data.

In the context of the study, proportions like 62% or 35% are used as samples from people with few or many social outlets, respectively.

• Each sample proportion provides an estimate of the true proportion in the entire population.
• This estimation helps in understanding and predicting the broader context of how different social behaviors might influence the occurrence of colds.

Although we deal with samples, these calculations aim to make informed, evidence-based guesses about the larger group, reflecting the randomness and chance inherent in such estimates.