Question

6.34 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). (Schmidt, 2011)

	Autism	Typical Development	Total
No vitamin	111	70	181
Vitamin	143	159	302
Total	254	229	483

(a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the three months before pregnancy and autism.

Ho: p_vitamin = p_{no vitamin}
Ha: p_vitamin < p_{no vitamin}

Ho: p_vitamin = p_{no vitamin}
Ha: p_vitamin ? p_{no vitamin}

Ho: p_vitamin = p_{no vitamin}
Ha: p_vitamin > p_{no vitamin}

(b) Complete the hypothesis test and state an appropriate conclusion.
What is the value of the test statistic for this test? (Please round to two decimal places) What is the p-value associated with this test? (Please round to four decimal places) The result of the hypothesis is:

Since p<? we fail to reject the null hypothesis

Since p ? ? we do not have enough evidence to reject the null hypothesis

Since p ? ? we reject the null hypothesis and accept the alternative

Since p ? ? we accept the null hypothesis

Since p<? we reject the null hypothesis and accept the alternative

Answer 1

a)Null Hypothesis Ho: pvitamin = pno vitamin
Alternative Hypothesis Ha: pvitamin ? pno vitamin

b) Under H0, the expected frequencies are

	Obs. Freq			Exp. Freq
	Autism	Typical Dev	Total	Autism	Typical Dev	Total
No Vitamin	113	70	181	95.18	85.82	181.00
Vitamin	143	159	302	158.82	143.18	302.00
Total	254	229	483	254	229	483.00

The calculation for test statistic is

	Obs. Freq( O )	Exp. Freq ( E )	(O-E)^2/E
	111	95.18	2.629
	70	85.82	2.916
	143	158.82	1.576
	159	143.18	1.748
Total	483	483	8.869

The test statistic is

Degrees of freedom = (2-1)*(2-1) = 1

The critical value of Chi square for 1 df at 5% significance level is 3.841

The P-Value is 0.0029

Since p<? we reject the null hypothesis and accept the alternative