A group of researchers studying the link between prenatal vitamin use and autism

Question

A group of 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 summarizes the number of mothers in each group who did and did not use prenatal vitamins during the three months before pregnancy (periconceptional period) (R.J. Schmidt et al. "Prenatal vitamins, one- carbon metabolism gene variants, and risk for autism". In: Epidemiology 22.4 (2011), p. 476.) Autism utism Typical development Total 70 159 229 Periconceptional No vitami prenatal vitamin Vitamin 143 254 181 302 483 Total (a) State appropriate hypotheses for testing whether there is a difference in the proportion of mothers (of children with and without autism) who took prenatal vitamins three months before pregnancy (b) Complete the hypothesis test manually and state an appropriate conclusion at the 5% level of significance. Make sure to verify any necessary conditions for the test. (c) Manually calculate a 95% confidence interval for the difference in the rate of using prenatal vitamins between mothers of children with autism and mothers of children without autism. (d) Does your confidence interval agree with the result of your hypothesis test? Must it agree? (e) Use Minitab or MS Excel to obtain the results in parts (b) and (c) for both pooled and unpooled version. Do you observe a difference in the final conclusion? Provide the corresponding Minitab or MS Excel output. (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 (R.C. Rabin. "Patterns: Prenatal Vitamins May Ward Off Autism". In: New York Times (2011))

Explanation / Answer

Part a

Here, we have to use z test for population proportions. Required null and alternative hypotheses are given as below:

Null hypothesis: H0: There is no any significant difference in the proportion of mothers who took vitamins three months before pregnancy.

Alternative hypothesis: Ha: There is a significant difference in the proportion of mothers who took vitamins three months before pregnancy.

H0: p1 = p2 versus Ha: p1 p2

This is a two tailed test.

Part b

H0: p1 = p2 versus Ha: p1 p2

Test statistic formula is given as below:

Z = (P1 – P2) /sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

We are given

Level of significance = = 0.05 or 5%

N1 = 254

N2 = 229

X1 = 143

X2 = 159

P1 = 143/254 = 0.562992126

P2 = 159/229 = 0.694323144

Z = (P1 – P2) /sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Z = (0.562992126 – 0.694323144) /sqrt[(0.562992126*(1 – 0.562992126)/254) + (0.694323144*(1 – 0.694323144)/229)]

Z = -2.9774

Critical value = -1.96 and 1.96

P-value = 0.0029

(By using z-table)

P-value < = 0.05

So, we reject the null hypothesis that there is no any significant difference in the proportion of mothers who took vitamins three months before pregnancy.

There is sufficient evidence to conclude that there is a significant difference in the proportion of mothers who took vitamins three months before pregnancy.

Part c

Formula for confidence interval for difference between two population proportions is given as below:

Confidence interval = (P1 – P2) -/+ Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

Confidence level = 95%

Critical z-value = 1.96 (by using z-table)

Confidence interval = (0.562992126 – 0.694323144) -/+ 1.96*sqrt[(0.562992126*(1 – 0.562992126)/254) + (0.694323144*(1 – 0.694323144)/229)]

Confidence interval = (0.562992126 – 0.694323144) -/+ 0.0853

Confidence interval = -0.13133102-/+ 0.0853

Lower limit = -0.13133102 - 0.0853 = -0.2167

Upper limit = -0.13133102 + 0.0853 = -0.0460

Part d

Yes, confidence interval is agree with the result of hypothesis test because the value zero is not lies between the given confidence interval.

Part e

Required Excel output is given as below:

Z Test for Differences in Two Proportions

Data

Hypothesized Difference

0

Level of Significance

0.05

Group 1

Number of Items of Interest

143

Sample Size

254

Group 2

Number of Items of Interest

159

Sample Size

229

Intermediate Calculations

Group 1 Proportion

0.562992126

Group 2 Proportion

0.694323144

Difference in Two Proportions

-0.13133102

Average Proportion

0.6253

Z Test Statistic

-2.9774

Two-Tail Test

Lower Critical Value

-1.9600

Upper Critical Value

1.9600

p-Value

0.0029

Reject the null hypothesis

Confidence Interval Estimate

of the Difference Between Two Proportions

Data

Confidence Level

95%

Intermediate Calculations

Z Value

-1.9600

Std. Error of the Diff. between two Proportions

0.0435

Interval Half Width

0.0853

Confidence Interval

Interval Lower Limit

-0.2167

Interval Upper Limit

-0.0460

Z Test for Differences in Two Proportions

Data

Hypothesized Difference

0

Level of Significance

0.05

Group 1

Number of Items of Interest

143

Sample Size

254

Group 2

Number of Items of Interest

159

Sample Size

229

Intermediate Calculations

Group 1 Proportion

0.562992126

Group 2 Proportion

0.694323144

Difference in Two Proportions

-0.13133102

Average Proportion

0.6253

Z Test Statistic

-2.9774

Two-Tail Test

Lower Critical Value

-1.9600

Upper Critical Value

1.9600

p-Value

0.0029

Reject the null hypothesis

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