The forward differences for this data is 1, 1, 1, 1, 1, 1 (since 10 - 9 = 1, 11 - 10 = 1, etc). Since we only need one iteration of differences, a linear polynomial will fit the data exactly.
Answer:
Since the null hypothesis is true, finding the significance is a type I error.
The probability of the year I error = level of significance = 0.05.
so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5
Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.
The pages would be 166 and 167
