(3)(10) A random sample of the birth weights of 186 babies has a mean of 3103g a
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(3)(10) A random sample of the birth weights of 186 babies has a mean of 3103g and a standard deviation of 696g (based on data from "Cognitive Outcomes of Preschool Children with Prenatal Cocaine Exposure," by Singer et al., Journal of the American Medical Association, Vol. 291, No. 20. These babies were born to mothers who did not use cocaine during their pregnancies. Further, a random sample of the birth weights of 190 babies born to mothers who used cocaine during their pregnancies has a mean of 2700g and a standard deviation of 645g. a) The birth weights of babies are known normally distributed. Use a 0.025 significance level to test the claim that both samples are from populations having the same standard deviation. b) Using your finding in (a), construct a 95% confidence interval estimate of the difference between the mean birth weight of a baby born to mothers who did not use cocaine and that of a baby born to mothers who used cocaine during their pregnancies. c Using your finding in (b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.Explanation / Answer
a.
Given that,
mean(x)=3103
standard deviation , s.d1=696
number(n1)=186
y(mean)=2700
standard deviation, s.d2 =645
number(n2)=190
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.025
from standard normal table, two tailed t /2 =2.25
since our test is two-tailed
reject Ho, if to < -2.25 OR if to > 2.25
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (185*484416 + 189*416025) / (376- 2 )
s^2 = 449854.7727
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=3103-2700/sqrt((449854.7727( 1 /186+ 1/190 ))
to=403/69.1826
to=5.8252
| to | =5.8252
critical value
the value of |t | with (n1+n2-2) i.e 374 d.f is 2.25
we got |to| = 5.8252 & | t | = 2.25
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 5.8252 ) = 0
hence value of p0.025 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 5.8252
critical value: -2.25 , 2.25
decision: reject Ho
p-value: 0
b.
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 3103-2700) ± t a/2 * sqrt( 449854.773 * (1/186+1/190) ]
= [ (403) ± 136.013 ]
= [266.987 , 539.013]
c.
1. we are 95% sure that the interval [266.987 , 539.013]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
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