Question

a) A data is collected from Lab A. Sample mean is 12, SD is 2.4 and sample size if 16. Another set of data from Lab B have a mean of 10 (assuming SD_B is also 2.4 and n_B=16). If we choose α = 0.05, do you think mean of Lab B is close enough to be considered the “same” as that of Lab A? Why?

b) Same as problem 4), except SD of Lab B is SD_B=4.4. If we choose α = 0.05, do you think the mean of Lab B is close enough to be considered the “same” as that of Lab A? Why?

Answer 1

(a)

Data:        

n1 = 16       

n2 = 16       

x1-bar = 12       

x2-bar = 10       

s1 = 2.4       

s2 = 2.4       

Hypotheses:        

Ho: μ1 = μ2        

Ha: μ1 ≠ μ2        

Decision Rule:        

α = 0.05       

Degrees of freedom = 16 + 16 - 2 = 30      

Lower Critical t- score = -2.042272449       

Upper Critical t- score = 2.042272449       

Reject Ho if |t| > 2.042272449       

Test Statistic:        

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   √(((16 - 1) * 2.4^2 + (16 - 1) * 2.4^2)/(16 + 16 -2)) =     2.4

SE = s * √{(1 /n1) + (1 /n2)} = 2.4 * √((1/16) + (1/16)) = 0.848528137      

t = (x1-bar -x2-bar)/SE = 2.357022604       

p- value = 0.025143982       

Decision (in terms of the hypotheses):        

Since 2.357022604 > 2.042272449 we reject Ho and accept Ha    

Conclusion (in terms of the problem):        

There is sufficient evidence that Lab B is significantly different from Lab A

(b)

Data:        

n1 = 16       

n2 = 16       

x1-bar = 12       

x2-bar = 10       

s1 = 2.4       

s2 = 4.4       

Hypotheses:        

Ho: μ1 = μ2        

Ha: μ1 ≠ μ2        

Decision Rule:        

α = 0.05       

Degrees of freedom = 16 + 16 - 2 = 30      

Lower Critical t- score = -2.042272449       

Upper Critical t- score = 2.042272449       

Reject Ho if |t| > 2.042272449       

Test Statistic:        

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   √(((16 - 1) * 2.4^2 + (16 - 1) * 4.4^2)/(16 + 16 -2)) =     3.544

SE = s * √{(1 /n1) + (1 /n2)} = 3.54400902933387 * √((1/16) + (1/16)) = 1.252996409      

t = (x1-bar -x2-bar)/SE = 1.596173769       

p- value = 0.120931822       

Decision (in terms of the hypotheses):        

Since 1.596173769 < 2.042272449 we fail to reject Ho    

Conclusion (in terms of the problem):        

There is no sufficient evidence that Lab B is significantly different from Lab A.