In buildings water leakage in an underground area facilitates condensation and i

Question

In buildings water leakage in an underground area facilitates condensation and it is very difficult to be repaired. To prevent the damages that could be caused by this leakage, an engineering studied the effects of different waterproofing methods in underground areas based on the changes of humidity to avoid condensation. Random samples of 20 and 18 specimens of waterproofing materials were applied to area into exterior and interior waterproofing construction methods and the changes in the humidity inside the specimens were observed. The data is given in following table. a) Using 5% significance level, test the equality of the two population variances. b) At the 5% significance level, can you conclude that mean humidity in the interior waterproofing is lower than that in the exterior waterproofing? c) Construct a 95% confidence interval for the difference of the two means. d) What are the assumptions needed to run the test in part (a)?

Explanation / Answer

(a)

Data:       

n1 = 20      

n2 = 18      

s1^2 = 2.25      

s2^2 = 2.56      

Hypotheses:      

Ho: 1^2 = 2^2      

Ha: 1^2 2^2      

Decision Rule:      

= 0.05      

Numerator DOF = 20 - 1 = 19    

Denominator DOF = 18 - 1 = 17    

Lower Critical F- score = 0.3896    

Upper Critical F- score = 2.6331    

Reject Ho if F < 0.389561 or F > 2.6331   

Test Statistic:      

F = s1^2 / s2^2 = 2.25/2.56 =   0.8789   

p- value = 0.609746      

Decision (in terms of the hypotheses):    

Since 0.389561 < 0.8789 < 2.6331 we fail to reject Ho

Conclusion (in terms of the problem):    

There is no sufficient evidence that the population variances are different

(b)

Data:      

n1 = 20     

n2 = 18     

x1-bar = 9     

x2-bar = 15.5     

s1 = 1.5     

s2 = 1.6     

Hypotheses:      

Ho: 1 2      

Ha: 1 < 2      

Decision Rule:      

= 0.05     

Degrees of freedom =   20 + 18 - 2 = 36   

Critical t- score =   -1.688297694    

Reject Ho if t <   -1.688297694    

Test Statistic:      

Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =      (((20 - 1) * 1.5^2 + (18 - 1) * 1.6^2) / (20 + 18 - 2)) = 1.548027419

SE = s * {(1 /n1) + (1 /n2)} = 1.54802741864894 * ((1/20) + (1/18)) = 0.502943496    

t = (x1-bar -x2-bar)/SE = (9 - 15.5)/0.502943496323222 = -12.923917    

p- value = 2.17344E-15     

Decision (in terms of the hypotheses):      

Since -12.923917 < -1.688297694 we reject Ho and accept Ha

Conclusion (in terms of the problem):    

There is sufficient evidence that 1 < 2

(c)

n1 = 20

n2 = 18

x1-bar = 9

x2-bar = 15.5

s1 = 1.5

s2 = 1.6

% = 95

Degrees of freedom = n1 + n2 - 2 = 20 + 18 -2 = 36

Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((20 - 1) * 1.5^2 + ( 18 - 1) * 1.6^2)/(20 + 18 -2)) = 1.548027419

SE = Pooled s * ((1/n1) + (1/n2)) = 1.54802741864894 * ((1/20) + (1/18)) = 0.502943496

t- score = 2.028093987

Width of the confidence interval = t * SE = 2.02809398678268 * 0.502943496323222 = 1.020016681

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = -6.5 - 1.02001668058458 = -7.520016681

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = -6.5 + 1.02001668058458 = -5.479983319

The 95% confidence interval is [-7.52, -5.48]

(d) The samples are assumed to be drawn from normally distributed populations.

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