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A screening program for neuroblastoma (a type of cancer) was undertaken in Germany among children born between November \(1,1993,\) and June \(30,2000,\) who were 9 to 18 months of age between May 1995 and April 2000 [6]. A total of 1,475,773 children participated in the screening program, of whom 204 were diagnosed between 12 and 60 months of age. The researchers expected the incidence rate of neuroblastoma to be 7.3 per 100,000 children during this period in the absence of screening. Test whether the number of cases detected by the screening program is significantly greater than expected. Provide a one-tailed \(p\) -value. (Hint: Use the normal approximation to the binomial distribution.)

Short Answer

Expert verified
The number of detected cases (204) is significantly greater than expected (\( p < 0.05 \)).

Step by step solution

01

Determine Expected Cases without Screening

Calculate the expected number of cases without screening using the expected incidence rate of 7.3 per 100,000 children for the given population size of 1,475,773 children. The expected number is determined by \( E = \frac{7.3}{100,000} \times 1,475,773 \approx 107.7359 \).
02

Set Up Hypothesis Test

Define the null hypothesis \( H_0 \) as the number of cases detected being equal to the expected number (107.7359), and the alternative hypothesis \( H_1 \) as the number of detected cases being greater than expected (204 > 107.7359).
03

Use Normal Approximation

Since the number of cases is large enough, approximate the binomial distribution using a normal distribution. Calculate the mean as \( \mu = E = 107.7359 \) and variance as \( \sigma^2 = E \). Hence, \( \sigma = \sqrt{107.7359} \approx 10.3795 \).
04

Calculate the Z-Score

Calculate the Z-score to determine how many standard deviations the observed number (204 cases) is above the expected number. The Z-score is given by \( Z = \frac{204 - 107.7359}{10.3795} \approx 9.288 \).
05

Calculate the One-Tailed p-value

Find the one-tailed \( p \)-value using the Z-score. For \( Z = 9.288 \), the \( p \)-value is the probability of observing a value greater than 204. Using standard normal distribution tables or a calculator, this \( p \)-value is approximately close to 0.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used to decide if a claim about a population is reasonable based on sample data. In this scenario, the claim is regarding the effectiveness of a neuroblastoma screening program in detecting more cases than expected. The core components of hypothesis testing include the null and alternative hypotheses.
  • Null Hypothesis (\(H_0\)): This assumes that there is no effect or difference. In our example, it means that the number of neuroblastoma cases detected is what was expected without the screening (107.7359 cases).
  • Alternative Hypothesis (\(H_1\)): This suggests that there is an effect or a difference. Here, it posits that the screening detected more cases than what would be expected without it (204 cases > 107.7359 cases).
Effective hypothesis testing involves rejecting or failing to reject the \(H_0\) based on calculated statistics and their comparison against critical values. A key outcome of this test is the \(p\)-value, which helps determine the statistical significance of results.
Normal Approximation
Normal approximation is a technique that simplifies calculations by approximating a complex distribution, like the binomial distribution, with a normal distribution under certain conditions. This is particularly useful when dealing with large sample sizes, where calculating exact binomial probabilities becomes cumbersome.In our problem, since the number of children involved in the screening program (1,475,773) is very large, normal approximation can be effectively used.
  • Mean (\(\mu\)): This is equivalent to the expected number of cases without screening, calculated as 107.7359.
  • Variance (\(\sigma^2\)): In normal approximation, the variance is also the expected number, giving us 107.7359, and the standard deviation (\(\sigma\)) becomes the square root of variance.
The approximate normal distribution of the number of cases detected can then be compared against reality to determine if the screening has had a notable impact.
Incidence Rate
Incidence rate helps us understand how frequently a condition, like neuroblastoma, occurs in a population over a specific period. It is often expressed per population size for easy comparisons across different contexts. In our example, this rate was expected to be 7.3 per 100,000 children. Knowing the incidence rate allows researchers to calculate the expected number of cases given the population size. It's crucial for establishing the baseline in hypothesis testing when judging the impact of a preventative measure like a health screening. By incorporating the incidence rate with the actual number of cases, experts can determine if the observed rate in the screened population is significantly different, supporting or refuting the effectiveness of the screening program.
Z-Score
Z-score is a statistical measure that describes how a data point compares to the mean of a data set, expressed in terms of standard deviations. It tells us how far away we are from the mean, and in which direction. In this context, the Z-score was used to compare the observed number of neuroblastoma cases (204) with the expected number (107.7359) under normal approximation.The Z-score is calculated as:\[Z = \frac{\text{Observed value} - \text{Expected mean}}{\text{Standard deviation}}\]For the screening, \[Z = \frac{204 - 107.7359}{10.3795} \approx 9.288\]This high Z-score indicates that the observed value is significantly higher than what was expected, suggesting a meaningful impact of the screening program on detecting neuroblastoma cases. It helps in deriving the \(p\)-value, providing insight into the statistical significance of this observation.

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