Solution to F/t problem

Question 1

Ho: s1 = s2

Ha: s1 ≠ s2

Numerator df = 24, denominator df = 16; reject Ho if F > 2.63 (note that left c.v. < 1 and not needed)

F = (24.2/18.1)² = 1.7876

Do not reject Ho.

Conclude no significant difference in the variability of the number of people walking by the two locations.

Question 2

Choose location 2 as group 1.

Ho: µ1 - µ2 ≤ 15

Ha: µ1 - µ2 > 15

Sp² = [(24(24.2)² + 16(18.1)²]/40 = 482.428

t = [(174.8 – 141.2) – 15]/sqrt(482.428/25 + 482.428/17) = 2.6938

The 1% critical value is 2.423 and the 0.5% critical value is 2.704 based on 40 d.f. Therefore,

0.5% < p-value < 1%. Since the p-value is less than 1%, reject Ho. Conclude the average number of people walking by location 1 is at least 15 less than the number of people walking by location 2.