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 Sample Mea Sample SD 26 19 808 790 115 72 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% CL with 99% confidence, we can say that the true average stance duration among elderly individuals falls outside these values. 0 with 99% confidence, we can say that the true average stance duration among elderly individuals falls below the lower bound. With 99% confidence, we can say that the true average stance duration among elderly individuals falls between these values. with 99% confidence, we can say that the true average stance duration among elderly individuals falls above the upper bound (b Carry out test of hypotheses at significance level 0.05 to decide whether true individuals.) State the relevant hypotheses. erage stance duration s larger mong elderly ndividuals than among younger individuals. Use 1 for elderly ndividuals and 2 for younger 0 H0: 1-2 = 0 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) -value State the conclusion in the problem context O Reject Ho. The data does not suggest that the true average stance duration is larger among elderly individuals than among younger individuals. Fail to reject Ho. The data does not suggest that the true average stance duration is larger among elderly individuals than among younger individuals Fail to reject Ho. The data suggests that the true average stance duration is larger among elderly individuals than among younger individuals. O Reject Ho. The data suggests that the true average stance duration is larger among elderly individuals than among younger individuals. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
mean(x)=808
standard deviation , s.d1=115
number(n1)=26
y(mean)=790
standard deviation, s.d2 =72
number(n2)=19
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((13225/26)+(5184/19))
= 27.955
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 18 d.f is 2.878
margin of error = 2.878 * 27.955
= 80.455
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (808-790) ± 80.455 ]
= [-62.455 , 98.455]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=808
standard deviation , s.d1=115
sample size, n1=26
y(mean)=790
standard deviation, s.d2 =72
sample size,n2 =19
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 = [( 808-790) ± t a/2 * sqrt((13225/26)+(5184/19)]
= [ (18) ± t a/2 * 27.955]
= [-62.455 , 98.455]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-62.455 , 98.455] 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 i sthe answer

PART B.
Given that,
mean(x)=808
standard deviation , s.d1=115
number(n1)=26
y(mean)=790
standard deviation, s.d2 =72
number(n2)=19
null, Ho: u1 - u2 = 0
alternate, H1: u1 - u2 > 0
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.552
since our test is right-tailed
reject Ho, if to > 2.552
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =808-790/sqrt((13225/26)+(5184/19))
to =0.6439
| to | =0.6439
critical value
the value of |t | with min (n1-1, n2-1) i.e 18 d.f is 2.552
we got |to| = 0.64389 & | t | = 2.552
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.6439 ) = 0.26388
hence value of p0.01 < 0.26388,here we do not reject Ho
ANSWERS
---------------
Ho: u1 - u2 = 0, H1: u1 - u2 > 0
test statistic: 0.64
p-value: 0.264
critical value: 2.552
Option B. decision: do not reject Hofailed to reject Ho, the data does not suggest that the true average
stance duration is narger among elder than youngers

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