91Ó°ÊÓ

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.

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."

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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.