Question

1. A gerentologist is studying whether Medicare patients receive the same attention from doctors as senior citizens with other insurance. She decides to measure this by average length of a preventative care visit. In the data she collects, 44 Medicare patients saw the doctor for a mean time of 18.72 minutes, with a standard deviation of 5.16 minutes; while 33 seniors with other insurance had a mean visit of 212 minutes, with a standard deviation of 6.17 minutes.
   1. Under the assumption of equal (population) variances, find a 98% confidence interval for the difference in visit length between the underlying groups.
   2. Given that population variances are not assumed equal, determine the degrees of freedom for the precise model, and find a 98% confidence interval for the difference in visit length.
   3. What conclusion should she reach based on each of these confidence intervals?

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=18.72
standard deviation , s.d1=5.16
number(n1)=44
y(mean)=21.2
standard deviation, s.d2 =6.17
number(n2)=33
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (43*26.626 + 32*38.069) / (77- 2 )
s^2 = 31.508
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 31.508 * (1/44+1/33) )
=1.293
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.02
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 75 d.f is 2.377
margin of error = 2.377 * 1.293
= 3.073
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (18.72-21.2) ± 3.073 ]
= [-5.553 , 0.593]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=18.72
standard deviation , s.d1=5.16
sample size, n1=44
y(mean)=21.2
standard deviation, s.d2 =6.17
sample size,n2 =33
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 = [( 18.72-21.2) ± t a/2 * sqrt( 31.508 * (1/44+1/33) ]
= [ (-2.48) ± 3.073 ]
= [-5.553 , 0.593]
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interpretations:
1. we are 98% sure that the interval [-5.553 , 0.593]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion

b.
TRADITIONAL METHOD
given that,
mean(x)=18.72
standard deviation , s.d1=5.16
number(n1)=44
y(mean)=21.2
standard deviation, s.d2 =6.17
number(n2)=33
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((26.626/44)+(38.069/33))
= 1.326
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.02
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 32 d.f is 2.449
margin of error = 2.449 * 1.326
= 3.248
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (18.72-21.2) ± 3.248 ]
= [-5.728 , 0.768]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=18.72
standard deviation , s.d1=5.16
sample size, n1=44
y(mean)=21.2
standard deviation, s.d2 =6.17
sample size,n2 =33
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 = [( 18.72-21.2) ± t a/2 * sqrt((26.626/44)+(38.069/33)]
= [ (-2.48) ± t a/2 * 1.326]
= [-5.728 , 0.768]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [-5.728 , 0.768] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion

c.
part (a) 98% sure that the interval [-5.553 , 0.593] with equal variances.
part (b) 98% sure that the interval [-5.728 , 0.768] with unequal variances.
slight change in the confidence interval.