Bocote is a type of wood desired for its aromatic properties, its beauty, and it
ID: 3227959 • Letter: B
Question
Bocote is a type of wood desired for its aromatic properties, its beauty, and its strength. A particular manufacturer has found it can recreate the aromatic properties and most of its aesthetic qualities, but not its strength, with a synthetic material. For a particular project, you can justify using (the more expensive) Bocote wood if its strength is more than 5 units on average stronger than the synthetic material. If you test 15 Bocote boards and find their strength to average 17 units with a standard deviation of 4.8 units and you test 22 synthetic boards to find an average of 14.2 and standard deviation of 2.8, is there enough evidence to justify using Bocote?
carry out the appropriate hypothesis test. Be sure to label all steps and draw real world conclusions
Explanation / Answer
Step 1: State the hypothesis
Null hypothesis = H0 = 1 - 2 <= 5
Alternative hypothesis = Ha = 1 - 2 > 5
Here 1 is the mean for Bocote wood and 2 is the mean for synthetic material
Step 2: Formulate an analysis plan - Lets take significane level of 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Step 3: Analyse sample data.
We need to calculate the standard error, degrees of freedom and t statistic
SE = sqrt[(s12/n1) + (s22/n2)]
s1 = 4.8, n1 = 15, s2 = 2.8, n2 = 22
SE = sqrt[ 4.8^2/15 + 2.8^2/22 ] = sqrt[ 1.536 + 0.3563636] = 1.375632
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
= (4.8^2/15 + 2.8^2/22)2 / { [ (4.8^2/15)2 / (15 - 1) ] + [ (2.8^2/22)2 / (22 - 1) ] }
= ( 1.536 + 0.3563636 )2 / { (1.5362 / 14) + (0.35636362 / 21) }
= 1.8923642 / { 0.1685211 + 0.006047382 }
= 3.581042 / 0.1745685 = 20.51368
t = [ (x1 - x2) - d ] / SE
= [(17 - 14.2) - 5] / 1.375632
= -2.2/1.375632
= -1.599265
p value associated with t statistic of -1.599265 is 0.0627.
i.e. P(T < -1.599265) = 0.0627. So P(T > -1.599265) = 1 - 0.0627 = 0.9373
So as p value is greater than the significance level of 0.05, we fail to reject the null hypothesis.
So there is not enough evidence to justify using Bocote.