Mathematics, 21.05.2020 04:05, mosesbalan7159

Autism and Maternal Antidepressant Use A recent study41 compared 298 children with Autism Spectrum Disorder to 1507 randomly selected con- trol children without the disorder. Of the children with autism, 20 of the mothers had used antidepres- sant drugs during the year before pregnancy or the first trimester of pregnancy. Of the control children, 50 of the mothers had used the drugs.
(a) Is there a significant association between prena- tal exposure to antidepressant medicine and the risk of autism? Test whether the results are sig- nificant at the 5% level.
(b) Can we conclude that prenatal exposure to antidepressant medicine increases the risk of autism in the child? Why or why not?

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(a) We conclude that there is a significant association between prenatal exposure to antidepressant medicine and the risk of autism at 5% significance level.

(b) Yes, we can conclude that prenatal exposure to antidepressant medicine increases the risk of autism in the child.

Step-by-step explanation:

We are given that Autism and Maternal Antidepressant Use A recent study 41 compared 298 children with Autism Spectrum Disorder to 1507 randomly selected control children without the disorder.

Of the children with autism, 20 of the mothers had used antidepressant drugs. Of the control children, 50 of the mothers had used the drugs.

Let p_1 = proportion of mothers who had used drugs having Autism Spectrum Disorder in their children.

p_2 = proportion of mothers who had used drugs not having Autism Spectrum Disorder in their children.

(a) Null Hypothesis, H_0 : p_1=p_2 {means that there is no significant association between prenatal exposure to antidepressant medicine and the risk of autism}

Alternate Hypothesis, H_A : $p_1\neq p_2$ {means that there is a significant association between prenatal exposure to antidepressant medicine and the risk of autism}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of mothers who had used drugs having Autism Spectrum Disorder in their children = $\frac{20}{298}$ = 0.067

$\hat p_2$ = sample proportion of mothers who had used drugs not having Autism Spectrum Disorder in their children = $\frac{50}{1507}$ = 0.033

n_1 = sample of children with Autism Spectrum Disorder = 298

n_2 = sample of children without Autism Spectrum Disorder = 1507

So, the test statistics = $\frac{(0.067-0.033)-(0)}{\sqrt{\frac{0.067(1-0.067)}{298}+\frac{0.033(1-0.033)}{1507} } }$

= 2.281

The value of z test statistics is 2.281.

Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a significant association between prenatal exposure to antidepressant medicine and the risk of autism.

(b) Yes, we can conclude that prenatal exposure to antidepressant medicine increases the risk of autism in the child because if we reconsider our hypothesis for one-tailed test in which the null hypothesis states that the prenatal exposure to antidepressant medicine does not increases the risk of autism in the child.

For this hypothesis also, our test statistics will be more than the critical value of z and due to which we reject our null hypothesis and conclude that prenatal exposure to antidepressant medicine increases the risk of autism.

answered: Guest

This does not make sense

answered: Guest

where are the choices?