The degenerative disease osteoarthritis most frequently affects weight-bearing j

Question

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. An article presented the following summary data on stance duration (ms) for samples of both older and younger adults. Age Older Younger Sample Size 27 16 Sample Mean 809 790 Sample SD 116 75 Assume that both stance duration distributions are normal (a) Calculate a 99% CI for true average stance duration among elderly individuals. )ms Interpret your 99% CI. O with 99% confidence, we can say that the true average stance duration among elderly individuals falls below the lower bound. O with 99% confidence, we can say that the true average stance duration among elderly individuals falls outside these values. O with 99% confidence, we can say that the true average stance duration among elderly individuals falls between these values. O with 99% confidence, we can say that the true average stance duration among elderly individuals falls above the upper bound (b) Carry out a test of hypotheses at significance level 0.05 to decide whether true average stance duration is larger among elderly individuals than among younger individuals. (Use 1 for elderly individuals and 2 for younger individuals.) State the relevant hypotheses.

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
mean(x)=809
standard deviation , s.d1=116
number(n1)=27
y(mean)=790
standard deviation, s.d2 =75
number(n2)=16
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((13456/27)+(5625/16))
= 29.154
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.01
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 15 d.f is 2.947
margin of error = 2.947 * 29.154
= 85.916
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (809-790) ± 85.916 ]
= [-66.916 , 104.916]
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DIRECT METHOD
given that,
mean(x)=809
standard deviation , s.d1=116
sample size, n1=27
y(mean)=790
standard deviation, s.d2 =75
sample size,n2 =16
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 = [( 809-790) ± t a/2 * sqrt((13456/27)+(5625/16)]
= [ (19) ± t a/2 * 29.154]
= [-66.916 , 104.916]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-66.916 , 104.916] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

Option C.
With 99% confidence, we can say that the true average stance duration among elderly individuals falls between these values.

PART B.
Given that,
mean(x)=809
standard deviation , s.d1=116
number(n1)=27
y(mean)=790
standard deviation, s.d2 =75
number(n2)=16
null, Ho: u1 - u2 = 0
alternate, H1: u1 - u2 > 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.753
since our test is right-tailed
reject Ho, if to > 1.753
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =809-790/sqrt((13456/27)+(5625/16))
to =0.65
| to | =0.65
critical value
the value of |t | with min (n1-1, n2-1) i.e 15 d.f is 1.753
we got |to| = 0.65172 & | t | = 1.753
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.6517 ) = 0.26222
hence value of p0.05 < 0.26222,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 - u2 = 0
alternate, H1: u1 - u2 > 0
test statistic: 0.65
p-value: 0.262
Option D

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