Question

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 = 5 normal subjects was 1.76 mm, and the sample standard deviation was 0.55; for n = 11 CTS subjects, the sample mean and sample standard deviation were 2.56 and 0.84, 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.)
Required:
Calculate the test statistic and determine the P-value.

Answer 1

Answer

test statistic $t = -2.27$    

P-value $p- value = 0.0198$

Step-by-step explanation:

From the question we are told that

     The  first  sample  size is  $n_1 = 5$

     The first  sample mean is  $\= x_1 = 1.76 \ mm$

     The first  standard is  $s_1 = 0.55$

     The second is $sample \ size\ is\ n_2 = 11$

     The second sample mean is  $\= x_2 = 2.56$

     The second standard deviation is  $\sigma = 0.84$

      The level of significance is $\alpha = 0.01$

The  null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

The  alternative hypothesis is  $H_a: \mu_1 - \mu_2 < 0$

Generally the test statistics is mathematically represented as

           $t = \frac{ (\= x _1 - \= x_2) - ( \mu_1 - \mu_2)}{ \sqrt{\frac{s_1 ^2 }{ n_1 } + \frac{s_1 ^2 }{ n_1 } } }$

=>        $t = \frac{ (1.76 - 2.56 ) - 0 }{ \sqrt{\frac{0.55^2 }{ 5 } + \frac{0.84^2 }{11 } } }$

=>        $t = -2.27$      

Generally degree of freedom is mathematically represented as

          $df = n_1 + n_2 - 2$

=>       $df = 5 + 11 - 2$

=>       $df = 14$

From the student t distribution table the probability value to the left that corresponds to $t = -2.27$  at a degree of freedom of  $df = 14$ is  

        $p- value = 0.0198$