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Researchers studying the link between prenatal vitamin use and autism surveyed t

ID: 3247474 • Letter: R

Question

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) (a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the three months before pregnancy and autism. H_0: P_vitamin = P_no vitamin H_a: P_vitamin > P_no vitamin H_0: P_vitamin = P_no vitamin H_a: P_vitamin notequalto P_no vitamin H_0: P_vitamin = P_no vitamin H_a: P_vitamin

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2

Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.62526

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.04411

z = (p1 - p2) / SE

z = - 2.98

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -2.98 or greater than 2.98.

Thus, the P-value = 0.0014

Interpret results. Since the P-value (0.0014) is less than the significance level (0.05), we have to reject the null hypothesis.

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