Question

In a study of chromosomal anomalies observed in a randomly selected sample of 12001200 infertile men...

In a study of chromosomal anomalies observed in a randomly selected sample of 12001200 infertile men with either a zero or low sperm count, a team of researchers assigned the value of 11 for the presence of any chromosomal anomaly in the subject’s sperm and the value of 00 for the absence of all chromosomal anomalies in the subject’s sperm.

To compare the rates of chromosomal anomalies, the team performed a standard normal test, or ?z‑test, for the difference of the sample proportions between the zero and low sperm count groups.

Using the software of your choice and the pooled estimate of the population proportion, ?̂ p^, compute the standard normal test statistic, or ?z‑statistic, for the difference between the proportions of chromosomal anomalies observed in the zero and the low sperm count groups, (?̂ zero−?̂ low)(p^zero−p^low). Round the ?z‑statistic to two decimal places.

Data:

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Homework Answers

Answer #1

We will solve it using R

> #Import the dataset
> d=read.csv(file.choose())
> head(d)
zero low
1 0 0
2 1 0
3 0 0
4 0 0
5 0 0
6 0 0

> table(d$zero)

0 1
450 150
> table(d$low)

0 1
585 15
> prop.test(x=c(150,15),n=c(600,600))

   2-sample test for equality of
   proportions with continuity
   correction

data: c(150, 15) out of c(600, 600)
X-squared = 126.17, df = 1, p-value <
2.2e-16
alternative hypothesis: two.sided
95 percent confidence interval:
0.1865024 0.2634976
sample estimates:
prop 1 prop 2
0.250 0.025

#Note that, for 2 x 2 table, the standard chi-square test in chisq.test() is exactly equivalent to prop.test() but it works with data in matrix form

# conclusion : Since the p-value < 2.2e-16 which is less than 0.05 which indicates that the proportion of zero and the low sperm count groups differ significantly.

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