A study was performed among subjects with persistent sciatica (El Barzouhi et al

Question

A study was performed among subjects with persistent sciatica (El Barzouhi et al. [17]). Sciatica is pain that radiates down the leg and is often caused by a herniated disk in the lumbar region of the lower back. A randomized trial was conducted comparing an early surgery strategy vs. conservative care for an additional 6 months followed by surgery for patients who did not improve. Of the 267 patients eligible for the study, 131 were randomly assigned to undergo early surgery and 136 to prolonged conservative care. Of the 131 patients in the surgery group, 15 recovered before surgery could be performed. Of the 136 patients in the conservative care group, 54 actually underwent surgery in the first year. Thus, a total of (131 15) + 54 = 170 patients actually received surgery and 15 + (136 54) = 97 patients received no surgery. The primary results were presented based on treatment actually received. An MRI was performed at 1 year of follow-up to determine whether disk herniation was present and, if so, the extent of herniation compared to baseline. Definite disk herniation was present for 15 patients who received surgery and for 25 patients who received no surgery. 10.53 What test can be performed to compare the prevalence of disk herniation between the patients who received surgery vs. the patients who did not receive surgery? 10.54 Perform the test in Problem 10.53, and report a twosided p-value. 10.55 Interpret the results in Problem 10.54 in words

Explanation / Answer

Given that,
sample one, x1 =170, n1 =267, p1= x1/n1=0.637
sample two, x2 =97, n2 =267, p2= x2/n2=0.363
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.637-0.363)/sqrt((0.5*0.5(1/267+1/267))
zo =6.318
| zo | =6.318
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =6.318 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 6.318 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 6.318
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support that to compare the prevalence of disk herniation between the patients who received surgery vs. the patients who did not receive surgery

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