Question

# 6.34 Prenatal vitamins and Autism: Researchers studying the link between prenatal vitamin use and autism surveyed...

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: pvitamin = pno vitamin
Ha: pvitamin < pno vitamin

Ho: pvitamin = pno vitamin
Ha: pvitamin ? pno vitamin

Ho: pvitamin = pno vitamin
Ha: pvitamin > pno 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

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

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