Question

Suppose that the rate of major congenital malformations in the general population is 2.5 malformations per 100 deliveries. (a) A study is set up to investigate if the offspring of Vietnam-veteran fathers are at special risk of having congenital malformations. 100 infants are identified in a birth registry as being offspring of Vietnam-veteran fathers and 4 have a major congenital malformation. We say there is an excess risk of malformations in this group if the probability of observing 4 or more major congenital malformations in the sample of size 100 is less than 0.05. Is there evidence to believe that there is an excess risk of malformations in this group? Show your calculations. (b) Using the same birth-registry data, the effect of maternal use of marijuana on the rate of major congenital malformations is also examined. Of 75 offspring of mothers who used marijuana, 8 are found to have a major congenital malformation. In this case, we say there is an excess risk of malformations in this group if the probability of observing 8 or more major congenital malformations in the sample of size 75 is less than 0.05. Is there evidence to believe that there is an excess risk of malformations in this group? Show your calculations. 5. Because serum cholesterol is related to age and sex, some investigators prefer to express it in terms of z-scores. If X is raw serum cholesterol, then z = X−µ σ , where µ is the mean and σ is the standard deviation of serum cholesterol for a given age-sex group. Suppose z is regarded as having a standard normal distribution. Further suppose that a person is regarded as having high cholesterol if z > 2. and borderline cholesterol if 1.5 < z < 2.0. (a) What proportion of people have high cholesterol? (b) What proportion of people have borderline cholesterol? (c) Suppose a person was told that his cholesterol level was in the top 10%. What is the minimum cholesterol level of that person?

Answer 1

(a)

Probability of having congenital malformations = 2.5 / 100 = 0.025

Probability of  observing 4 or more major congenital malformations in the sample of size 100 is,

= P(X 4) where X ~ Bin(n = 100, p = 0.025)

= 1 - P(X < 4)

= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) ]

= 1 - (¹⁰⁰C₀ * 0.025⁰ * (1 - 0.025)¹⁰⁰ + ¹⁰⁰C₁ * 0.025¹ * (1 - 0.025)⁹⁹ + ¹⁰⁰C₂ * 0.025² * (1 - 0.025)⁹⁸ + ¹⁰⁰C₃ * 0.025³ * (1 - 0.025)⁹⁷ ]

= 1 - (0.07951729 + 0.20389049 + 0.25878408 + 0.21675931)

= 0.2410488

Since the probability of  observing 4 or more major congenital malformations in the sample of size 100 is not less than 0.05. there is no sufficient evidence to believe that there is an excess risk of malformations in this group.

(b)

Probability of congenital malformation = 8 / 75 = 0.1067

Probability of observing 8 or more major congenital malformations in the sample of size 75 is less than 0.05

= P(X 8)

Using Normal approximation of Binomial distribution, X ~ N( = 8 , = = 2.67)

= P[Z (8 - 8) / 2.67]

= P[Z 0]

= 0.5

Since the probability of observing 8 or more major congenital malformations in the sample of size 75 is not less than 0.05, there is no sufficient evidence to believe that there is an excess risk of malformations in this group.

5.

(a)

Proportion of people have high cholesterol = P(z > 2) = 0.0228 (Using Z tables)

(b)

Proportion of people have borderline cholesterol = P(1.5 < z < 2.0) = P(z < 2.0) - P(z < 1.5)

= 0.9772 - 0.9332 (Using z tables)

= 0.044

(c)

For the top 10%, the percentile is 90%. Z value for 90% percentile is 1.28

Minimum cholesterol level of that person will be have z value = 1.28