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# Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel...

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). An article reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for m = 7 normal subjects was 1.85 mm, and the sample standard deviation was 0.57; for n = 11 CTS subjects, the sample mean and sample standard deviation were 2.47 and 0.85, respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of 0.01. (Use μ1 for normal subjects and μ2 for CTS subjects.)

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.)

 t = P-value =

## Solutions

##### Expert Solution

Given that,
mean(x)=1.85
standard deviation , s.d1=0.57
number(n1)=7
y(mean)=2.47
standard deviation, s.d2 =0.85
number(n2)=11
null, Ho: u1 = u2
alternate, H1: u1 ≠ u2
level of significance, alpha = 0.01
from standard normal table, two tailed t alpha/2 =3.71
since our test is two-tailed
reject Ho, if to < -3.71 OR if to > 3.71
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.85-2.47/sqrt((0.3249/7)+(0.7225/11))
to =-1.8518
| to | =1.8518
critical value
the value of |t alpha| with min (n1-1, n2-1) i.e 6 d.f is 3.71
we got |to| = 1.85181 & | t alpha | = 3.71
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p ≠ -1.8518 ) = 0.114
hence value of p0.01 < 0.114,here we do not reject Ho
---------------
null, Ho: u1 = u2
alternate, H1: u1 ≠ u2
test statistic: -1.8518
critical value: -3.71 , 3.71
decision: do not reject Ho
p-value: 0.114

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