: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.

= av. of sample differences / SE of av

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

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

= " " " 38/6 = 6.3

= 2.5

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

SE of sum of differences = 2.7 x square root of 6

= 2.7 x 2.45 = 6.6

SE of average of differences = 6.6/6 = 1.1

t = 3/1.1 = 2.72

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%".