BONUS
CHAPTER
4

Solutions to “Practice Problems”

Bonus Chapter 1

1. 

 Since F > F_c, we reject H₀ and conclude that there is a difference between the sample means.

 F_sc = (k – 1)F_{α, k – 1, N – k} = (3 – 1)(3.739) = 7.478

 We conclude that there is a difference between gas mileage of Cars 1 and 2 and Cars 1 and 3.

4. 

 Since F’ > F_C’, we reject H₀’ and conclude that the blocking procedure was effective and proceed to test H₀.

 Since F > F_C, we reject H₀ and conclude that there is a difference between the golfer means.

Bonus Chapter 2

1. H₀ : The arrival process can be described by the expected distribution

 H₁ : The arrival process differs from the expected distribution

 For α = 0.05 and d.f. = k – 1 = 6 – 1 = 5, . Since , we do not reject H₀ and conclude that the arrival distribution is consistent with the expected distribution.

2. H₀ : The process can be described with the Poisson distribution using λ = 3.

 H₁ : The process differs from the Poisson distribution using λ = 3.

 For α = 0.01 and d.f. = k – 1 = 8 – 1 = 7, . Since , we do not reject H₀ and conclude that the process is consistent with the Poisson distribution using λ = 3.

3. H₀ : Grades are independent of reading time

 H₁ : Grades are dependent of reading time

 Sample expected frequency calculations:

 For α = 0.05 and d.f. = (r – 1)(c – 1) = (3 – 1)(5 – 1) = 8, . Since , we reject H₀ and conclude that there is a relationship between grades and the number of hours reading.

4. H₀ : The process can be described with the Binomial distribution using p = 0.4.

 H₁ : The process differs from the Binomial distribution using p = 0.4.

Because we must meet the chi-square requirement of having at least five observations in each of the expected frequency categories, we combined the final two categories in the following table (4 and 5 visits).

 For α = 0.05 and d.f. = k – 1 = 5 – 1 = 4, . Since , we reject H₀ and conclude that the process differs from the Binomial distribution using p = 0.4.

 Since t > t_c, we reject H₀ and conclude the correlation coefficient is not equal to zero.

 2a.

2b. H₀ : β = 0, H₁ : β ≠ 0

 d.f. = n – 2 = 10 – 2 = 8, t_c = 2.306

 Since t > t_c, we reject H₀ and conclude there is a relationship between payroll and wins.

 2c. = 51.21 + 0.294(70) = 71.79

 2d.

 CI = 71.79 ± (3.355)(10.26)(0.325) = 71.79 ± 11.19, (60.60, 82.98)

 2e. R² = (0.782)² = 0.612 or 61.2 percent

 3. GMAT GPA