:6. Solution. The SE of the first weighing is 3 ounces, that of
the second weighing 4 ounces. The SE of the 1st weighing minus
the second is given by the square root of 4^2 +3^2, or 5. (This
is because the error made by the second scale is independent of
that made by the first scale.) The null hypothesis is that the
true weight of the cheese was the same for both weighings. Under
the null hypothesis, Z, the observed difference minus the
expected difference divided by the SE of the differences, is
equal to
((15 -10) - 0)/5 = 1

The histograms of the possible first weighings and second
weighings both follow the normal curve, hence the histogram of
the possible differences between the two weighings also follows
the normal curve. It follows that the chance that Z is bigger
than the observed value of 1 is approximately equal to the area
under the normal curve to the right of 1. The p-value is close to
(100-68)/2 = 32/2 = 16%. This is closest to 15%.