14. Seat Belts A study of seat belt use involved children who were hospitalized

Question

14. Seat Belts A study of seat belt use involved children who were hospitalized atter vehicle crashes. For a group of 123 children who were wearing seat belts, the number of days in intensive care units (ICU) has a mean of 0.83 and a standard deviation of 1.77. For a gro of 290 children who were not wearing seat belts, the number of days spent in of 1.39 and a standard deviation of 3.06 (based on data from “Morbidity Among Pediatric Mo. tor Vehicle Crash Victims: The Effectiveness of Seat Belts." by Osberg and Di Scala, American Journal of Public Health, Vol. 82, No. 3). up ICUs has a mean a. Use a 0.05 significance level to test the claim that children wearing seat belts have a lower mean length of time in an ICU than the mean for children not wearing seat belts. b. Construct a confidence interval appropriate for the hypothesis test in part (a). c. What important conclusion do the results suggest?

Explanation / Answer

PART A.
Given that,
mean(x)=0.83
standard deviation , s.d1=1.77
number(n1)=123
y(mean)=1.39
standard deviation, s.d2 =3.06
number(n2)=290
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.657
since our test is left-tailed
reject Ho, if to < -1.657
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.83-1.39/sqrt((3.1329/123)+(9.3636/290))
to =-2.3301
| to | =2.3301
critical value
the value of |t | with min (n1-1, n2-1) i.e 122 d.f is 1.657
we got |to| = 2.33012 & | t | = 1.657
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -2.3301 ) = 0.01072
hence value of p0.05 > 0.01072,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.3301
critical value: -1.657
decision: reject Ho
p-value: 0.01072

PART B.
TRADITIONAL METHOD
given that,
mean(x)=0.83
standard deviation , s.d1=1.77
number(n1)=123
y(mean)=1.39
standard deviation, s.d2 =3.06
number(n2)=290
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((3.133/123)+(9.364/290))
= 0.24
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 122 d.f is 1.98
margin of error = 1.98 * 0.24
= 0.476
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (0.83-1.39) ± 0.476 ]
= [-1.036 , -0.084]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=0.83
standard deviation , s.d1=1.77
sample size, n1=123
y(mean)=1.39
standard deviation, s.d2 =3.06
sample size,n2 =290
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 0.83-1.39) ± t a/2 * sqrt((3.133/123)+(9.364/290)]
= [ (-0.56) ± t a/2 * 0.24]
= [-1.036 , -0.084]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.036 , -0.084] 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

PART C.
we support the claim, children wearing seat belts have a lower mean length
of a time

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