Question

At BurgerJoint, consistency in product and service is the new motto. Accordingly, management has invested in an automated french-fry dispenser to replace the manual dispensing method. The goal is to standardize the number of fries provided. The accompanying table shows the number of fries dispensed in a large-sized container based on samples of 10 orders before and after the process change. Assume these samples were drawn randomly and independently from normally distributed populations. (You may find it useful to reference the appropriate table: chi-square table or F table)

Manual	27	39	29	31	30	37	41	29	38	31
Automated	30	35	37	38	34	35	35	32	35	32

a. Select the hypotheses to test whether the variance of the new automated dispensing method is lower than the previous manual method.

• H₀: σ₁² / σ₂² = 1, H_A: σ₁² / σ₂² ≠ 1

• H₀: σ₁² / σ₂² ≤ 1, H_A: σ₁² / σ₂² > 1

• H₀: σ₁² / σ₂² ≥ 1, H_A: σ₁² / σ₂² < 1

b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

b-2. Find the p-value.

• p-value < 0.01

• 0.01 ≤ p-value < 0.025

• 0.025 ≤ p-value < 0.05

• 0.05 ≤ p-value < 0.10

• p-value ≥ 0.10

c. What is the conclusion at the 1% significance level?

Answer 1

a)

H₀: σ₁² / σ₂² ≥ 1, H_A: σ₁² / σ₂² < 1

b-1)

test statsitic =(s1/s2)² = 0.231

for one tailed hypothesis:

use excel function f.test(1st array,2nd array)/2 =0.020

0.01 ≤ p-value < 0.025

c)

since p value is not less than 0.01 , we can not reject the null hypothesis