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Chapter 20: Problem 33

A new vaccine was recently tested to see if it could prevent the painful and recurrent ear infections that many infants suffer from. The Lancet, a medical journal, reported a study in which babies about a year old were randomly divided into two groups. One group received vaccinations; the other did not. During the following year, only 333 of 2455 vaccinated children had ear infections, compared to 499 of 2452 unvaccinated children in the control group. a. Are the conditions for inference satisfied? b. Find a \(95 \%\) confidence interval for the difference in rates of ear infection. c. Use your confidence interval to explain whether you think the vaccine is effective.

Short Answer

Expert verified
The rate of ear infection in vaccinated children is between \(3.34 \%\) and \(10.18 \%\) lesser than in unvaccinated children with \(95 \%\) confidence, indicating that the vaccine may effectively reduce ear infections in infants.

Step by step solution

01

Check Conditions For Inference

There are two main conditions for making inferences about a population proportion. The first is random sampling, which is satisfied here as the babies were randomly divided into two groups. The second is that the sample size should be sufficiently large, which we can check by ensuring that \(np \geq 10\) and \(n(1-p) \geq 10\). Plugging in the numbers we get for vaccinated group \(2455 * (333/2455) = 333 \geq 10\) and \(2455 * (1 - 333/2455) = 2122 \geq 10\). For unvaccinated group \(2452 * (499/2452) = 499 \geq 10\) and \(2452 * (1 - 499/2452) = 1953 \geq 10\). So, the condition is satisfied.
02

Calculate Point Estimate and Standard Error

The point estimate for the difference in proportions is \(p_1 - p_2 = (333/2455) - (499/2452) = -0.0676\). The standard error for this estimate is \(\sqrt{[p_1(1-p_1)/n_1] + [p_2(1-p_2)/n_2]} = \sqrt{[(333/2455)(1-333/2455)/2455] + [(499/2452)(1-499/2452)/2452]} = 0.0174\).
03

Calculate Confidence Interval

A \(95 \%\) confidence interval for the difference in proportions is given by the formula \((p_1-p_2) \pm z*SE\), where \(z = 1.96\) for \(95 \%\) confidence. Plug the values from previous calculations to get \(-0.0676 \pm 1.96*0.0174\), which gives \(-0.1018, -0.0334\) as the confidence interval.
04

Interpret Results

With \(95 \%\) confidence, the proportion of vaccinated children getting ear infections is between \(3.34 \%\) and \(10.18 \%\) less than the proportion of unvaccinated children getting ear infections.

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Key Concepts

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

Vaccination Effectiveness
Understanding the effectiveness of a vaccine is key in public health. Here, we are exploring whether a new vaccine can lower the incidence of ear infections in infants.
We can gauge the effectiveness of the vaccine by observing the difference in infection rates between vaccinated and unvaccinated groups.
This examination was conducted using a statistical method called 'difference in proportions'. A vaccine is considered effective if the vaccinated group displays a significantly lower proportion of infections compared to the unvaccinated group. In our study, only 333 out of 2455 vaccinated children got ear infections, whereas 499 out of 2452 unvaccinated children experienced them.
When we convert these into proportions, this gives 13.6% for the vaccinated group and 20.3% for the unvaccinated group. This suggests that the vaccine could lower the risk of infections significantly. Effectiveness isn't just about percentages, though. It also involves statistical tools, such as confidence intervals, to determine how reliably these results suggest a real difference that isn't just due to chance.
Statistical Inference
Statistical inference allows us to make conclusions about a population based on a sample. Here, inference helps us determine if the new vaccine effectively reduces ear infection rates.
For reliable statistical inference, certain conditions must be met, such as random sampling and adequate sample size. In this exercise, babies were randomly divided into two groups鈥攖hose vaccinated and those not vaccinated.
This randomization satisfies one condition for inference, minimizing the risk of bias. The second condition concerns the sample size, which must be large enough to make valid predictions.
In our case, both groups exceeded the size threshold, meaning that statistical tests are less likely to give inaccurate results. Therefore, we can use statistical inference to confidently assert that the observed differences in infection rates are not accidental.
Difference in Proportions
The difference in proportions is a measure used to examine the effectiveness of treatment or intervention between two groups.
In this study, we want to find out the difference in proportions between the vaccinated and unvaccinated groups. This difference can tell us how effective the vaccination is in preventing infections.First, we calculate the point estimate of the difference. The formula is simple: \[ p_1 - p_2 \]where \( p_1 \) and \( p_2 \) are the infection rates for the vaccinated and unvaccinated groups, respectively.
For this study, the point estimate is -0.0676, indicating a lower infection rate in the vaccinated group. To understand how precisely this point estimate reflects the true difference, we construct a confidence interval. A 95% confidence interval, for example, would show us a range in which we are 95% certain the true difference in proportions lies. This methodical approach allows us to back up claims about vaccine effectiveness with statistical evidence, thus assuring us that the results are robust and not merely due to random chance. In this scenario, the negative interval indicates the vaccinated group indeed has fewer infections.

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Most popular questions from this chapter

Suppose an advocacy organization surveys 960 Canadians and 192 of them reported being born in another country (www.unitednorthamerica.org/simdiff.htm). Similarly, 170 out of 1250 U.S. citizens reported being foreign-born. Find the standard error of the difference in sample proportions.

Data collected in 2015 by the Behavioral Risk Factor Surveillance System revealed that in the state of New Jersey, \(27.3 \%\) of whites and \(47.2 \%\) of blacks were cigarette smokers. Suppose these proportions were based on samples of 3607 whites and 485 blacks. a. Create a \(90 \%\) confidence interval for the difference in the percentage of smokers between black and white adults in New Jersey. b. Does this survey indicate a race-based difference in smoking among American adults? Explain, using your confidence interval to test an appropriate hypothesis. c. What alpha level did your test use?

A study published in the Archives of General Psychiatry examined the impact of depression on a patient's ability to survive cardiac disease. Researchers identified 450 people with cardiac disease, evaluated them for depression, and followed the group for 4 years. Of the 361 patients with no depression, 67 died. Of the 89 patients with minor or major depression, 26 died. Among people who suffer from cardiac disease, are depressed patients more likely to die than non-depressed ones? a. What kind of design was used to collect these data? b. Write appropriate hypotheses. c. Are the assumptions and conditions necessary for inference satisfied? d. Test the hypothesis and state your conclusion. e. Explain in this context what your \(\mathrm{P}\) -value means. \(f\). If your conclusion is actually incorrect, which type of error did you commit? g. Create a 95\% Cl. h. Interpret your interval. i. Carefully explain what "95\% confidence" means.

In Exercises 49 and 50 , Chapter 4 , we looked at data from an experiment to determine whether visual information about an image helped people "see" the image in \(3 D\). 2-Sample \(t\) -Interval for \(\mu 1-\mu 2\) Conf level \(=90 \%\) df \(=70\) \(\mu(\mathrm{NV})-\mu(\mathrm{VV})\) interval : (0.55,5.47) a. Interpret your interval in context. b. Does it appear that viewing a picture of the image helps people "see" the \(3 \mathrm{D}\) image in a stereogram? c. What's the margin of error for this interval? d. Explain carefully what the \(90 \%\) confidence level means. e. Would you expect a \(99 \%\) confidence interval to be wider or narrower? Explain. f. Might that change your conclusion in part b? Explain.

When a random sample of 935 parents were asked about rules in their homes, \(77 \%\) said they had rules about the kinds of TV shows their children could watch. Among the 790 of those parents whose teenage children had Internet access, \(85 \%\) had rules about the kinds of Internet sites their teens could visit. That looks like a difference, but can we tell? Explain why a two-sample \(z\) -test would not be appropriate here.
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