:7. Solution.  The sample differences are 6, 1, 3, -1, 6, 3 . 
Since we assume that the sample comes from a population whose
histogram follows the normal curve, and since we don't know
anything about the SD of the population, we can and should use
the t-test.
    t = (av.of the sample differences - EV of av.)/SE of av. <p>

= av. of sample differences / SE of av <p>

Av of sample differences = (6+1+3-1+6+3)/6 = 18/6 = 3  <p>

Sample SD = square root of (9+4+0+16+9+0)/6  <p>

=   "     "    " 38/6 = 6.3  <p>

= 2.5  <p>

SD+ =  2.5 x square root of 6/5 = 2.5 x 1.1 = 2.7  <p>

SE of sum of differences = 2.7 x square root of 6  <p>

= 2.7 x 2.45 
 	
 
 	
 = 6.6  <p>

SE of average of differences = 6.6/6 = 1.1  <p>

t =  3/1.1 = 2.72  <p>

Since df = 6-1 = 5, the p-value is the area under the t-curve
with 5 degrees of freedom to the right of 2.72. Looking at the
t-table, we see that 95% of the area under the t-curve with df=5
lies between -2.57 and 2.57. 5/2 = 2.5% of the area is to the
right of 2.57.  Thus the p-value is less than 2.5%.  Since the
observed value of z is less than 4.02, the p-value is bigger that
.05%.  The right answer is "bigger than .05% but less than or
equal to 2.5%".