Question

The family planning wing of the health department of a certain state wishes to conduct a survey at a university campus for estimating the average time gap between the births of children in families having two children. The frame available, of course, lists all the 400 families of the campus. As the prior identification of the families in the population, who have just two children was difficult, the investigator selected a WOR random sample of 50 families. In the sampled families, 20 families were found having two children. These 20 families were interviewed, and the information collected was:

Family	1	2	3	4	5	6	7	8	9	10
Gap	54	34	44	44	31	26	51	52	44	59
Family	11	12	13	14	15	16	17	18	19	20
Gap	30	33	57	29	26	42	60	34	31	49

Estimate the average gap between the births of two children, and obtain confidence limits for it.

Answer 1

Let X be the variable that denote 'Gap'. Average of n values X1, X2,...,Xn is given by:

Mean=Average=830/20=41.5

Standard deviation is given by

These values can be found out using R-Software as follows.

R-commands and outputs:

Gap=c(54,34,44,44,31,26,51,52,44,59,30,33,57,29,26,42,60,34,31,49)
Xbar=mean(Gap)
Xbar
[1] 41.5

Original data is assumed to be Normal. As the standard deviation is to be estimated and sample size is small (n=20), t-distribution is used.
100(1-alpha)% confidence interval is [Xbar-tc*s/sqrt(n),Xbar-tc*s/sqrt(n)]
where,

• tc=critical value of t at alpha level with (n-1) degrees of freedom,
• Xbar=sample mean,
• s=sample standard deviation,
• n=sample size

# sd() command gives the standard deviation
s=sd(Gap)
s
[1] 11.45931

n=length(Gap)
n
[1] 20

Generally, 95% confidence interval is found out.
alpha=0.05
# qt() command gives the quantile for t-distribution
# Critical value can also be found using Statistical Table.
tc=qt(1-alpha/2,df=20-1)
tc
[1] 2.093024

Xbar-tc*s/sqrt(n)
[1] 36.13688
Xbar+tc*s/sqrt(n)
[1] 46.86312

# 95% Confidence limits/interval is [36.13, 46.86].