: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%.