Question

Two brands of microwave popcorn are compared for consistency in popping time.  The popping times, in seconds, for 30 randomly selected bags of each brand are provided in the attached file ‘Popcorn’.  Use the technology of your choice to answer the following questions.

a. We are interested in determining whether the standard deviations of the popping times differ for the two brands of popcorn. Conduct an appropriate analysis of the data values and determine if the assumptions of the F-Test procedure for testing the equality of variances are satisfied.

b. Conduct an F-test for the equality of the standard deviations of the two brands. Are the standard deviations different at the 0.01 significance level?

Construct a 99% confidence interval for the ratio of the standard deviations. Which of the following statements is true?

		The confidence interval includes the value zero (0) which indicates there is no difference in the standard deviations.
		The confidence interval includes the value zero (0) which indicates there is a difference in the standard deviations.
		The confidence interval includes the value one (1) which indicates there is no difference in the standard deviations.
		The confidence interval includes the value one (1) which indicates there is a difference in the standard deviations.

BRAND_A	BRAND_B
147	146
150	152
151	153
139	150
165	146
153	152
154	154
158	141
146	157
153	149
145	152
142	150
158	158
145	143
143	150
157	148
155	149
138	150
147	141
147	145
153	149
164	147
140	158
155	143
151	158
144	154
146	145
151	149
160	152
147	158

Answer 1

a) Assumptions:

Population is approximately normal.

Samples are independent and random.

b) Brand A, Sample 1:  
Sample standard deviation, s₁ = STDEV.S() = 7.0110
n₁ =    30
Brand B, Sample 2:  
s₂ =    4.9861
n₂ =    30
α =    0.01
  
Null and alternative hypothesis:  
Hₒ : σ₁ = σ₂  
H₁ : σ₁ ≠ σ₂  
  
Test statistic:  
F = s₁² / s₂² = 1.9772
  
Degree of freedom:  
df₁ = n₁-1 =    29
df₂ = n₂-1 =    29
  
Critical value(s):  
Lower tailed critical value, F_α/2 = F.INV(0.005, 29,29) =  0.3740
Upper tailed critical value, F_1-α/2 = F.INV(0.995, 29,29) =   2.6740
  
P-value :  
P-value = 2*F.DIST.RT(1.9772, 29, 29) =   0.0714

Decision: Do not reject the null hypothesis.

99% confidence interval for standard deviation:

Lower Bound = SQRT((s₁² / s₂²)/F_₁-α/₂) = 0.8599
Upper Bound = SQRT((s₁² / s₂²)/F_α/₂) = 2.2993

The confidence interval includes the value one (1) which indicates there is no difference in the standard deviations.