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The article "Spray Flu Vaccine May Work Better Than Injections for Tots" (San Luis Obispo Tribunc. May 2, 2006 ) described a study that compared flu vaccine administered by injection and flu vaccine administered as a nasal spray. Each of the 8000 children under the age of 5 who participated in the study received both a nasal spray and an injection, but only one was the real vaccine and the other was salt water. At the end of the flu season, it was determined that \(3.9 \%\) of the 4000 children receiving the real vaccine by nasal spray got sick with the flu and \(8.6 \%\) of the 4000 recciving the real vaccine by injection got sick with the flu. a. Why would the researchers give every child both a nasal spray and an injection? b. Use the given data to estimate the difference in the proportion of children who get sick with the flu after being vaccinated with an injection and the proportion of children who get sick with the flu after being vaccinated with the nasal spray using a \(99 \%\) confidence interval. Based on the confidence interval, would you conclude that the proportion of children who get the flu is different for the two vaccination methods?

Short Answer

Expert verified
a. Researchers gave every child both treatments to create a double-blind study, which aids in reducing bias. b. After calculating the confidence interval, if zero is not in the interval it can be concluded that there is a significant difference between the two vaccination methods.

Step by step solution

01

Understanding the Experimental Design

Every child received both a nasal spray and an injection to ensure that the children, their parents, and potentially the doctors administering the vaccines have no idea what real vaccine was given. This is called a double-blind study, which can reduce the bias. The salt water serves as a placebo.
02

Calculation of Sample Proportions

Calculate the sample proportions of children who got sick with the flu for each method of vaccination. For the nasal spray, \(3.9\%\) of the 4000 children which equals \(0.039 \times 4000 = 156\) children. This gives a sample proportion of \(p_1 = 156/4000 = 0.039\). Likewise, for the injection, \(8.6\%\) of the 4000 children or \(0.086 \times 4000 = 344\) children got sick with flu. This gives a sample proportion \(p_2 = 344/4000 = 0.086\).
03

Estimation of Standard Error

Use the formula for standard error \(SE\) of the difference in proportions which is \(SE = \sqrt{ \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} }\). Substituting \(p_1=0.039\), \(p_2=0.086\), and sample sizes \(n_1=n_2=4000\), we get the calculated value for the standard error.
04

Calculation of Confidence Interval

Use the following formula for a \(99\%\) confidence interval for the difference in proportions: \( (p_1 - p_2) \pm z*SE \). Here, \(z = 2.576\) for \(99\%\) confidence. Substituting the values will result in the confidence interval.
05

Interpretation of Confidence Interval

If zero is not in the confidence interval, then it can be concluded that there is a significant difference in the proportions of children getting sick with two vaccination methods.

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Key Concepts

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

Understanding Experimental Design in Vaccine Studies
In vaccine studies, experimental design is crucial to obtaining reliable results. The study discussed used a method called the double-blind experiment. This means that neither the participants (children and their parents) nor the administrators knew which vaccine was real. By giving each child both a nasal spray and an injection, the researchers aimed to prevent bias. This design ensures that the placebo effect doesn't influence the outcome. The saltwater solution used as a placebo further strengthens the study by creating identical conditions for both groups.
  • The real vaccine and placebo were indistinguishable.
  • Ensures that all children were treated similarly.
  • Helps in objectively assessing the vaccine's effectiveness.
Such designs are essential for maintaining integrity in medical research.
Exploring Confidence Intervals
A confidence interval is a range of values that estimates a certain parameter. In our case, it estimates the difference in flu symptoms between nasal spray and injection vaccines. With a 99% confidence interval, there's a high chance, 99% to be precise, that the interval contains the true difference in proportions. It offers a useful measure of reliability for our estimate.
  • A wide interval suggests more variability and less precision.
  • Narrow intervals indicate higher precision.
  • At 99%, our confidence is particularly strong, minimizing error.
Confidence intervals help us understand how likely our results are to represent the whole population accurately.
Comparing Proportions in Vaccine Efficacy
Proportion comparison is key in identifying the effectiveness of different vaccines. In this study, proportions help us quantify how many children got sick in each vaccine group. For the nasal spray group, 3.9% got sick, while the injection group had an 8.6% illness rate. The differences in these proportions suggest a potentially significant result.
  • Proportion for nasal spray: 0.039
  • Proportion for injection: 0.086
  • Difference indicates nasal spray may be more effective.
If a confidence interval for the difference excludes zero, it's statistically significant; the proportions aren't equal, suggesting one method might be superior in this case.

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Most popular questions from this chapter

"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006\()\). The article states that "those with a college degree reported a higher incidence of sunburn that those without a high school degree- \(43 \%\) versus \(25 \% .\) " For purposes of this exercise, suppose that these percentages were based on random samples of size 200 from cach of the two groups of interest (college graduates and those without a high school degrec). Is there convincing evidence that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degrec? Answer based on a test with a \(.05\) significance level.

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The director of the Kaiser Family Foundation's Program for the Study of Entertainment Media and Health said, "It's not just teenagers who are wired up and tuned in, its babies in diapers as well." A study by Kaiser Foundation provided one of the first looks at media use among the very youngest children- those from 6 months to 6 years of age (Kaiser Family Foundation. 2003 , www .kff.org). Because previous research indicated that children. who have a TV in their bedroom spend less time reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of parents. One sample consisted of parents of children age 6 months to 3 years old. The second sample consisted of parents of children age 3 to 6 years old. They found that the proportion of children who had a TV in their bedroom was \(.30\) for the sample of children age 6 months to 3 years and \(.43\) for the sample of children age 3 to 6 years old. Suppose that the two sample sizes were each 100 . a. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 6 months to 3 years who have a TV in their bedroom. Hint: This is a one-sample confidence interval. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 3 to 6 years who have a 'TV in their bedroom. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportion that have TVs in the bedroom for children age 6 months to 3 years and for children age 3 to 6 years. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

The article referenced in the previous exercise also reported that \(24 \%\) of the males and \(16 \%\) of the females in the 2006 sample reported owning an \(\mathrm{MP} 3\) player. Suppose that there were the same number of males and females in the sample of \(1112 .\) Do these data provide convincing evidence that the proportion of females that owned an MP3 player in 2006 is smaller than the corresponding proportion of males? Carry out a test using a significance level of 01 .

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