/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 A study was recently published i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A study was recently published in Western Australia on the relationship between method of conception and prevalence of major birth defects (Hansen et al. [8]). The prevalence of at least one major birth defect among infants conceived naturally was \(4.2 \%,\) based on a large sample of infants. Among 837 infants bom as a result of invitro fertilization (IVF), 75 had at least one major birth defect. Does an unusual number of infants have at least one birth defect in the IVF group? Why or why not? (Hint: Use an approximation to the binomial distribution.)

Short Answer

Expert verified
No, the number of birth defects in the IVF group is unusually high, as indicated by a very high Z-score (6.86).

Step by step solution

01

Understand the Problem

We are given the probability of a major birth defect in naturally conceived infants, which is 4.2%, and the data for 837 IVF infants with 75 having major birth defects. The task is to determine if the number of birth defects in the IVF group is unusual.
02

Define the Null Hypothesis

The null hypothesis assumes that the rate of major birth defects in IVF infants is the same as that of naturally conceived infants. Thus, under the null hypothesis, the probability of a major birth defect, \( p \), is 0.042.
03

Calculate Expected Number of Birth Defects

With IVF infants sample size \( n = 837 \), we calculate the expected number of birth defects using the formula \( np \). \[ np = 837 \times 0.042 = 35.154 \]
04

Determine Variance and Standard Deviation

The variance for a binomial distribution is \( np(1-p) \). Calculate the variance and the standard deviation: \[ \text{Variance} = 837 \times 0.042 \times (1-0.042) = 33.691 \] \[ \text{Standard Deviation} = \sqrt{33.691} \approx 5.803 \]
05

Calculate Z-score

To find how unusual it is to have 75 birth defects, calculate the Z-score: \[ Z = \frac{\text{Observed} - \text{Expected}}{\text{Standard Deviation}} = \frac{75 - 35.154}{5.803} \approx 6.86 \]
06

Interpret the Z-score

A Z-score of 6.86 is extremely high, indicating that observing 75 or more birth defects is highly unlikely under the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics, often used in biostatistics to draw conclusions from data. It's like a formal tool to check if the observed results differ significantly from what was expected. In our context, it helps to determine whether IVF births lead to more birth defects compared to natural conceptions.

Here's the basic structure:
  • **Null Hypothesis ( H_0 ):** This is the default hypothesis that states there is no effect or no difference. For this study, it means that IVF and natural conception results in the same rate of birth defects.
  • **Alternative Hypothesis ( H_a ):** This hypothesis suggests there is a difference. Here, it would mean the birth defect rate is higher in IVF births.
Hypothesis testing involves calculating a statistic from your data and then assessing this statistic's significance compared to a threshold (often called the alpha level). In this case, calculating the Z-score is part of this process to understand if 75 IVF birth defects are statistically unusual.
Binomial Distribution
The binomial distribution is crucial for any situation where there are only two possible outcomes for each trial, which we're examining in this study: whether an infant does or does not have a major birth defect.

This distribution is defined by two parameters:
  • **The number of trials (n):** Here, it's the total number of IVF infants studied, which is 837.
  • **The probability of success (p):** For naturally conceived infants, the probability of a birth defect is 0.042 (or 4.2%).
The expected number of birth defects in IVF infants can be calculated using the formula np, where n is the total number, and p is the probability. The variance and standard deviation also fall under this distribution, helping us measure variability around the expected outcomes. In this case, the standard deviation helps determine how far-off our observed value of 75 birth defects is from what we'd expect under the null hypothesis.
Birth Defects Study
A birth defects study focuses on understanding the prevalence and potential causes of congenital disabilities in newborns. In this context, it looks specifically at infants conceived via methods like IVF compared to natural conception.

These studies involve collecting data from a large group of infants to determine how frequently birth defects occur. They help us understand whether certain factors potentially elevate the risk. For this study:
  • Researchers found 4.2% birth defect prevalence in naturally conceived infants.
  • They observed 75 of 837 infants from IVF births had major birth defects.
The study's aim is to identify if IVF conception contributes to a higher instance of birth defects. This could have significant implications for healthcare practices and parental advice.
IVF vs Natural Conception
In comparing IVF (In Vitro Fertilization) to natural conception, researchers aim to determine if there are health differences in offspring resulting from the two methods. IVF is a process where an egg is fertilized by sperm outside the body and then implanted in the womb.

The key inquiries in studies involving these two methods of conception include:
  • Does the environment of conception - lab versus natural - affect infant health outcomes?
  • Are there specific risks associated with IVF that parents should consider?
By examining differences, especially in birth defect prevalence, the study from Western Australia seeks to uncover if IVF could be a contributing factor to higher birth defect rates. Their findings were significant enough to show that in their sample, IVF infants exhibited a statistically larger number of birth defects than those conceived naturally. Such insights can guide future research and influence fertility treatment practices.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Blood pressure readings are known to be highly variable. Suppose we have mean SBP for one individual over \(n\) visits with \(k\) readings per visit \(\left(\bar{X}_{n, k}\right) .\) The variability of \(\left(\bar{X}_{n, k}\right)\) depends on \(n\) and \(k\) and is given by the formula \(\sigma_{w}^{2}=\sigma_{A}^{2} / n+\) \(\sigma^{2} /(n k),\) where \(\sigma_{A}^{2}=\) between visit variability and \(\sigma^{2}=\) within visit variability. For \(30-\) to 49 -year-old Caucasian females, \(\sigma_{A}^{2}=42.9\) and \(\sigma^{2}=12.8 .\) For one individual, we also assume that \(\bar{X}_{n, k}\) is normally distributed about their true long- term mean \(=\mu\) with variance \(=\sigma_{w}^{2}\). Suppose a woman is measured at two visits with two readings per visit. If her true long-term SBP \(=130 \mathrm{mm} \mathrm{Hg}\) then what is the probability that her observed mean SBP is \(\geq 140 \mathrm{mm}\) Hg? (lgnore any continuity correction.) ( Note : By true mean SBP we mean the average SBP over a large number of visits for that subject.)

The Diabetes Prevention Trial (DPT) involved a weight loss trial in which half the subjects received an active intervention and the other half a control intervention. For subjects in the active intervention group, the average reduction in body mass index (BMI, i.e., weight in kg/height\(^2\) in \(\mathrm{m}^{2}\) ) over 24 months was \(1.9 \mathrm{kg} / \mathrm{m}^{2}\). The standard deviation of change in BMI was \(6.7 \mathrm{kg} / \mathrm{m}^{2}\). If the distribution of BMI change is approximately normal, then what is the probability that a subject in the active group would lose at least 1 BMI unit over 24 months?

Because serum cholesterol is related to age and sex, some investigators prefer to express it in terms of \(z\) -scores. If \(X=\) raw serum cholesterol, then $$Z=\frac{X-\mu}{\sigma}$$, where \(\mu\) is the mean and \(\sigma\) is the standard deviation of serum cholesterol for a given age-gender group. Suppose \(Z\) is regarded as a standard normal random variable. What is \(\operatorname{Pr}(-1.0

A doctor diagnoses a patient as hypertensive and prescribes an antihypertensive medication. To assess the clinical status of the patient, the doctor takes \(n\) replicate blood-pressure measurements before the patient starts the drug (baseline) and \(n\) replicate blood-pressure measurements 4 weeks after starting the drug (follow-up). She uses the average of the \(n\) replicates at baseline minus the average of the \(n\) replicates at follow-up to assess the clinical status of the patient. She knows, from previous clinical experience with the drug, that the mean diastolic blood pressure (DBP) change over a 4-week period over a large number of patients after starting the drug is \(5.0 \mathrm{mm}\) Hg with variance \(33 / n,\) where \(n\) is the number of replicate measures obtained at both baseline and follow-up. The physician also knows that if a patient is untreated (or does not take the prescribed medication), then the mean DBP over 4 weeks will decline by 2 mm Hg with variance 33/n. What is the probability that an untreated subject will decline by at least \(5 \mathrm{mm}\) Hg if 1 replicate measure is obtained at both baseline and follow-up?

It is also known that over a large number of 30 - to 49 -year-old Caucasian women, their true mean SBP is normally distributed with mean \(=120 \mathrm{mm}\) Hg and standard deviation \(=14\) mm Hg. Also, over a large number of African American 30- to 49-year-old women, their true mean SBP is normal with mean \(=130 \mathrm{mm} \mathrm{Hg}\) and standard deviation \(=20 \mathrm{mm} \mathrm{Hg}\). Suppose we select a random \(30-\) to 49 -year-old Caucasian woman and a random 30 - to 49 -year-old African American woman. What is the probability that the African American woman has a higher true SBP?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.