Question

summary data on proportional stress limits for specimens constructed using two different types of wood are shown below:

Type of Wood	Sample Size	Sample Mean	Sample St. Dev.
Red Oak	14	8.7	0.81
Douglas Fir	10	6.49	1.08

Assuming that both samples were selected from normal distributions with similar population variances, calculate a 99% confidence interval for the true mean difference in proportional stress limit of wood. The lower confidence interval bound for the two sided confidence interval is:

		0.7499
		1.4114
		1.0701
		1.1246
		3.2954

Answer 1

Solution:

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 8.7

X2bar = 6.49

S1 = 0.81

S2 = 1.08

n1 = 14

n2 = 10

df = n1 + n2 - 2 = 22

Confidence level = 99%

Critical t value = 2.8188

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(14 – 1)* 0.81^2 + (10 – 1)* 1.08^2]/(14 + 10 – 2)

S_p² = 0.8649

Confidence interval = (8.7 – 6.49) ± 2.8188*sqrt[0.8649*((1/14)+(1/10))]

Confidence interval = 2.21 ± 2.8188*0.3850

Confidence interval = 2.21 ± 1.085238

Lower limit = 2.21 - 1.085238 = 1.1246

Upper limit = 2.21 + 1.085238 = 3.2954

Confidence interval = (1.1246, 3.2954)

Answer: Lower bound = 1.1246