4. 3/4 points | Previous Answers DevoreStat9 9.E.021 My Notes Ask Your T Quantit

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4. 3/4 points | Previous Answers DevoreStat9 9.E.021 My Notes Ask Your T 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 1.78 mm, and the sample standard deviation was 0.53; for n 10 TS subjects, the sample mean and sample standard deviation were 2.55 and 0.81, 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. (UseA1 for normal subjects and 2 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= -2.37 p-value0.017 State the conclusion in the problem context. Fail to reject Ho. The data suggests that the true average gap detection threshold for CTS subjects is the same as that for normal subjects. O Reject Ho. The data suggests that the true average gap detection threshold for CTS subjects is the same as that for normal subjects. O Reject Ho. The data suggests that the true average gap detection threshold for CTS subjects exceeds that for normal subjects. O Fail to reject Ho. The data suggests that the true average gap detection threshold for CTS subjects exceeds that for normal subjects. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

Given that,
mean(x)=1.78
standard deviation , s.d1=0.53
number(n1)=7
y(mean)=2.55
standard deviation, s.d2 =0.81
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.707
since our test is two-tailed
reject Ho, if to < -3.707 OR if to > 3.707
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.78-2.55/sqrt((0.2809/7)+(0.6561/10))
to =-2.37
| to | =2.37
critical value
the value of |t | with min (n1-1, n2-1) i.e 6 d.f is 3.707
we got |to| = 2.36796 & | t | = 3.707
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.368 ) = 0.056
hence value of p0.01 < 0.056,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.37
critical value: -3.707 , 3.707
decision: do not reject Ho
ANSWERS:
p-value: 0.056
Conclusion: Fail to reject H0. the data suggests that the true average gap detection threshold for CTS subjcts is the same as that for normal subjects

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