One tailed test of hypothesis: difference between means

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To evaluate the relative merits of two prosthetic devices designed to facilitate manual dexterity, an occupational therapist assigned 21 patients with identical handicaps to wear one or the other of two devices while performing a certain task. 11 patients wore device A and 10 wore device B. The researcher recorded the time each patiend required to perform a certain task, with the following results

Device A Device B
mean= 65 seconds mean=75 seconds
variance= 81 seconds^2 variance= 64 seconds^2

Do these data provide sufficient evidence to indicate that device A is more effective than device B
The significance level is not stated in this problem. We will take it to be equal to
0.05 or 5%
To test the effectiveness of the devices we have the research (or Alternative hypothesis that mean of A > mean of B)

One tailed test for difference between means (small sample size)

At significance level= 0.05

A B
Mean =M = 65.00 75.00 Difference of means= 10.00 =absolute value of 65 - 75
s^2= 81.00 64.00
Standard deviation =s=square root of s^2 9 8
Sample size=n= 11 10

First we will calculate the pooled estimate of statndard deviation using standard deviations for A and B
pooled estimate of s^2=sp^2={ (n1-1)s1^2+(n2-1)s2^2}/(n1+n2-2)=
=(10 * 9 ^2 + 9 * 8 ^2 )/ 19 = 72.9474
pooled estimate of s=sp= = square root of 72.9474 = 8.54

Standard error of difference of mean =s x1-x2 =spsquare root of (1/n1+1/n2)=
=(8.54 * square root of ( 1/ 11 + 1/ 10 )= 3.73

Null Hypothesis: Ho: M A=M B There is no difference between the means
Alternative Hypothesis: H1: M A>M B There is difference between the means

degrees of freedom=n1+n2-2= 19

Significance level= 0.05

t-value= 1.7291 corresponding to level of significance= 0.05 and degrees of freedom= 19

We calcualte the t value using the excel function TINV

M x1-x2 = Hypothesized difference of mean= 0
Upper limit of acceptance region=M x1-x2+t*sx1-x2= 6.45 = 0 + 1.7291 * 3.73

Since the difference of means= 10.00 is greater than the upper limit of acceptance region= 6.45
Accept alternative hypothesis; The difference of means is not equal to 0

Therefore these data provide sufficient evidence at 0.05 level of significance
(or 95% confidence) that A is more effective than B

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