Question

The article "Spray Flu Vaccine May Work Better Than Injections for Tots" (San Luis Obispo Tribune, 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 receiving 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

There is a significant difference in the proportion of children who get the flu after being vaccinated with an injection and the proportion of children who get the flu after being vaccinated with the nasal spray. The 99% confidence interval for the difference in proportions is [0.0249, 0.0691]. As for the reason for giving each child both a nasal spray and an injection, this is a form of experimental control, aimed at accounting for any external factors that might influence the results, such as individual immune responses.

Step-by-step solution

Understand the problem and gather the data

The problem gives you the percentages of the children who got sick after receiving the real vaccine either by nasal spray or injection. Convert these percentages into proportions: \(3.9 \% \) = \(0.039 \) for the nasal spray and \(8.6 \% \) = \(0.086 \) for the injection. Since there were 4000 kids in each group, you can determine the number that got sick in each group by multiplying, giving \(4000*0.039 = 156 \) for spray and \(4000*0.086 = 344 \) for injection.

Calculate the difference in proportions

This can be done by subtracting the proportion of kids who got sick with the nasal spray from the proportion of kids who got sick with the injection. That is, \(0.086 - 0.039 = 0.047\).

Calculate the Standard Error

The standard error can be calculated using the formula for the standard error of the difference between two proportions: \(\sqrt{(p1*(1-p1)/n1) + (p2*(1-p2)/n2)}\), where p1 and p2 are the proportions in the two groups, and n1 and n2 are the sizes of the two groups. Substituting the given values yields SE ≈ 0.0093.

Calculate the Confidence Interval

The formula to calculate a 99% confidence interval is: \( p1 - p2 \pm Z(α/2) * SE\), where Z(α/2) is the critical value from the standard normal distribution corresponding to a two-tail test with α = 1 % (0.01). From the Z table, we find the critical value ≈ 2.33. So the confidence interval is \(0.047 \pm 2.33 * 0.0093 = [0.0249, 0.0691]\).

Conclusion

Since the confidence interval does not contain 0, there is a significant difference between the proportion of children who get the flu using the different vaccination methods.

Key concepts

Proportion Difference

Understanding proportion differences is crucial in many scientific studies, notably when comparing effectiveness between two groups or treatments. In the exercise, researchers aimed to compare the flu contraction rates between children receiving a flu vaccine via nasal spray versus an injection. The proportions, 3.9% got sick from the spray group and 8.6% from the injection group, highlight a proportion difference of 4.7 percentage points (or 0.047 as a decimal).

This numerical difference provides valuable initial insight into the efficacy of one method over the other. However, calculating this difference is merely the first step—interpreting this difference within a broader statistical context requires evaluating its variability and significance, hence the need for further steps such as standard error calculation and confidence interval determination.

Standard Error Calculation

The standard error (SE) measures the variance in sample statistics—here, the difference between the two proportions—from one sample to another. It is an estimate of the true sampling distribution's standard deviation. Calculating the SE in the context of a difference in proportions lays the foundation for constructing confidence intervals and making inferences about the desired population.

The formula used in the solution, \(\sqrt{(p1*(1-p1)/n1) + (p2*(1-p2)/n2)}\), combines the variances of two independent samples to estimate the SE of the difference. This calculated SE, approximately 0.0093, tells us that if similar studies were repeated under the same conditions, the observed difference in proportions would typically vary by about 0.0093 due to sampling variability alone.

Statistical Significance

Statistical significance helps to determine if the observed results are due to chance or to a specific factor—in this case, the method of vaccination. By constructing a 99% confidence interval, researchers are indicating that if the experiment were repeated many times, 99 out of 100 times the true difference in proportions would lie within this interval.

The confidence interval in the solution, [0.0249, 0.0691], doesn't contain 0, indicating that the observed 4.7 percentage point difference is likely not due to random chance. This suggests a statistically significant difference in flu rates based on the method of vaccination. In practical terms, it presents strong evidence in favor of the nasal spray's effectiveness in preventing flu compared to injections—for the study's age group and conditions.