/*! 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 34 Prenatal vitamins and Autism. Re... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 34

Prenatal vitamins and Autism. Researchers studying the link between prenatal vitamin use and autism surveyed the mothers of a random sample of children aged \(24-60\) months with autism and conducted another separate random sample for children with typical development. The table below shows the number of mothers in each group who did and did not use prenatal vitamins during the three months before pregnancy (periconceptional period). 40 \begin{tabular}{llccc} & \multicolumn{4}{c} { Autism } \\ \cline { 3 - 4 } & & Autism Typical development & Total \\ \cline { 2 - 5 } Periconceptional & No vitamin & 111 & 70 & 181 \\ prenatal vitamin & Vitamin & 143 & 159 & 302 \\ \cline { 2 - 5 } & Total & 254 & 229 & 483 \end{tabular} (a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the three months before pregnancy and autism. (b) Complete the hypothesis test and state an appropriate conclusion. (Reminder: verify any necessary conditions for the test.) (c) A New York Times article reporting on this study was titled "Prenatal Vitamins May Ward Off Autism", Do you find the title of this article to be appropriate? Explain your answer. Additionally, propose an alternative title.

Short Answer

Expert verified
The study suggests an association but does not prove causation. A more accurate article title would be "Study Suggests Association Between Prenatal Vitamins and Lower Autism Rates."

Step by step solution

01

Define the Hypotheses

To test for independence between prenatal vitamin use and autism, let's establish the null and alternative hypotheses. - Null Hypothesis \(H_0\): There is no association between the use of prenatal vitamins during the periconceptional period and autism in children.- Alternative Hypothesis \(H_a\): There is an association between the use of prenatal vitamins during the periconceptional period and autism in children.
02

Set Up the Contingency Table

Use the provided data to set up the contingency table for the chi-square test of independence.\[\begin{array}{|c|c|c|c|}\hline & \text{No Vitamin} & \text{Vitamin} & \text{Total} \\hline\text{Autism} & 111 & 143 & 254 \\hline\text{Typical Development} & 70 & 159 & 229 \\hline\text{Total} & 181 & 302 & 483 \\hline\end{array}\]
03

Calculate Expected Counts

Based on the numbers in the table, calculate the expected counts using the formula:\[E_{ij} = \frac{(\text{Row Total})_i \times (\text{Column Total})_j}{\text{Grand Total}}\]- Expected for "No Vitamin, Autism": \(E_{11} = \frac{254 \times 181}{483} \approx 95.21\)- Expected for "Vitamin, Autism": \(E_{12} = \frac{254 \times 302}{483} \approx 158.79\)- Expected for "No Vitamin, Typical": \(E_{21} = \frac{229 \times 181}{483} \approx 85.79\)- Expected for "Vitamin, Typical": \(E_{22} = \frac{229 \times 302}{483} \approx 143.21\)
04

Perform Chi-Square Test

Use the chi-square formula to calculate the test statistic:\[\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}\]Calculating:- \((111 - 95.21)^2 / 95.21 \approx 2.69\)- \((143 - 158.79)^2 / 158.79 \approx 1.67\)- \((70 - 85.79)^2 / 85.79 \approx 2.96\)- \((159 - 143.21)^2 / 143.21 \approx 1.75\)Sum all these values: \(\chi^2 \approx 9.07\). With 1 degree of freedom, because the table is a 2x2 contingency table, compare \(\chi^2\) value to the critical value (3.84 at \(\alpha = 0.05\)).
05

Conclusion of Hypothesis Test

Since the calculated chi-square statistic (9.07) is greater than the critical value (3.84), we reject the null hypothesis. Therefore, there is sufficient evidence to suggest an association between the periconceptional use of prenatal vitamins and autism.
06

Evaluate the Article Title

The title "Prenatal Vitamins May Ward Off Autism" suggests causation, which this study does not establish because it is only observational. A more appropriate title would be "Study Suggests Association Between Prenatal Vitamins and Lower Autism Rates."

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 91影视!

Key Concepts

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

Chi-Square Test
The Chi-Square Test is a statistical method used to determine if there is a significant association between two categorical variables. In this context, it helps to identify whether prenatal vitamin usage is related to autism in children. The process starts by setting up a contingency table, which organizes the data based on categories such as vitamin use and autism diagnosis. The next step involves calculating the expected counts for each cell within the table. These expected values are based on the assumption that there is no association between the variables. Then, the chi-square statistic is computed by comparing the observed data to these expected counts. Specifically, it quantifies how much the observed data deviate from what would be expected under independence. If the chi-square statistic is larger than the critical value from the chi-square distribution (based on the chosen significance level and degrees of freedom), we reject the null hypothesis.
Contingency Table
A Contingency Table is a matrix format used to display the frequency distribution of variables and is crucial for conducting a Chi-Square Test. In our exercise, it consists of data on prenatal vitamin use and autism incidence. The table is set up with rows and columns representing different categories of the variables. For instance, one category might be 'No Vitamin' versus 'Vitamin', and the other could be 'Autism' versus 'Typical Development'. This arrangement helps to neatly organize the data, allowing us to easily visualize and compute the expected counts for each combination of categories. Further, the table allows for the calculation of totals for each row and column, which are essential for determining expected values. Contingency tables provide a clear, comprehensive view of the relationships between the variables before further statistical analysis.
Null Hypothesis
The Null Hypothesis (\(H_0\)) is a foundational concept in hypothesis testing, asserting that there is no association or effect between the variables being studied. For this exercise, it states that prenatal vitamin use has no impact on the likelihood of autism development in children. The starting assumption for hypothesis testing is that the null hypothesis is true until proven otherwise. Thus, the chi-square test is employed to test this assumption. If the evidence from the test is strong enough, as indicated by the test statistic surpassing the critical value, the null hypothesis can be rejected. However, this does not prove the alternative hypothesis; it simply implies that the data collected are inconsistent with the null hypothesis, suggesting a potential association. The null hypothesis is crucial because it provides a benchmark against which the statistical evidence is measured.

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

Is college worth it? Part. I. Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, \(48 \%\) said they decided not to go to college because they could not afford school. 40 (a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement. (b) Would you expect a confidence interval for the proportion of American adults who decide not to go to college because they cannot afford it to include 0.5? Explain.

Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is \(8.0 \%\), while this proportion is \(8.8 \%\) for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a \(95 \%\) confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. \(^{44}\)

Open source textbook. A professor using an open source introductory statistics book predicts that \(60 \%\) of the students will purchase a hard copy of the book, \(25 \%\) will print it out from the web, and \(15 \%\) will read it online. At the end of the semester he asks his students to complete a survey where they indicate what format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30 said they printed it out from the web, and 25 said they read it online. (a) State the hypotheses for testing if the professor's predictions were inaccurate. (b) How many students did the professor expect to buy the book, print the book, and read the book exclusively online? (c) This is an appropriate setting for a chi-square test. List the conditions required for a test and verify they are satisfied. (d) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value. (e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpret your conclusion in this context.

Gender and color preference. A 2001 study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A \(95 \%\) confidence interval for the difference between the proportions of males and females whose favorite color is black ( \(p_{\text {male }}-p\) female \()\) was calculated to be (0.02,0.06) . Based on this information, determine if the following statements are true or false, and explain your reasoning for each statement you identify as false. 2 (a) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) lower to \(6 \%\) higher than the true proportion of females whose favorite color is black. (b) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) to \(6 \%\) higher than the true proportion of females whose favorite color is black. (c) \(95 \%\) of random samples will produce \(95 \%\) confidence intervals that include the true difference between the population proportions of males and females whose favorite color is black. (d) We can conclude that there is a significant difference between the proportions of males and females whose favorite color is black and that the difference between the two sample proportions is too large to plausibly be due to chance. (e) The \(95 \%\) confidence interval for \(\left(p_{\text {female }}-p_{\text {male }}\right)\) cannot be calculated with only the information given in this exercise.

True or false, Part. II. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(X^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation
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.