In vitro fertilization (IVF) treatment Box 4.7 highlights the study performed by

Question

In vitro fertilization (IVF) treatment Box 4.7 highlights the study performed by Malizia et al. (2009) to determine the cumulative live-birth rates resulting from IVF treatment over a course of six cycles. In Box 4.7, a comparison of population proportion of live births for two different age groups was performed when an optimistic approach was adopted to account for the outcomes of non-returnees in the cohort. Here, we look for differences in population proportions of live births within the age groups when a conservative approach is used, i.e. all non-returnees to subsequent IVF cycles (i.e. cycles 2 to 6) are assumed to have failed outcomes.

The conservative cumulative probabilities for live births observed at the end of each IVF cycle are displayed in Table P4.1. The cohort is divided into four groups based on maternal age. The cohort of 6164 women is categorized according to age as in Table P4.2.

Determine at the end of the sixth IVF cycle whether the probabilities of live births for the age groups of less than 35 years of age and equal to or more than 35 years of age are significantly different.

Explanation / Answer

Solution:

For the given cohort study, we have to compare the population proportion of live births for two different age groups. For this comparison purpose we have to use the z test for population proportions. First of all, we have to find out the sample proportions for the given two groups.

For the group age ‘less than 35’ years

X = 2678

For the group ‘equal to or more than 35’ years of age

X = 1360 + 836 + 1290 = 3486

N = 2678 + 1360 + 836 + 1290 =6164

First sample proportion = P1 = 2678/6164 = 0.434458, Q1 = 1 - 0.434458 = 0.565542

Second sample proportion = P2 = 3486/6164 = 0.565542, Q2 = 1 – 0.565542 = 0.434458

H0: Two population proportions are same.

Ha: Two population proportions are not same.

Test statistic formula is given as below:

Z = (P1 – P2) / sqrt[(P1Q1/N1) + (P2Q2/N2)]

We are given N1 = N2 = N = 6164

Z = (0.434458 - 0.565542) / sqrt((0.434458*0.565542/6164)+( 0.565542*0.434458/6164))

Z = -14.6811

P-value = 0.0000 (approximately)

Here, p-value is very less.

So, we reject the null hypothesis two population proportions are same.

We conclude that there is sufficient evidence that two population proportions are not same.

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