Answer:
The calculated value of t= -0.21 does not lie in the critical region t= ± 2.120 Therefore we accept our null hypothesis that there's no difference between strength (psi) of liner specimens when a certain fusion process was used and when this process was not used at the 0.05 significance level
Step-by-step explanation:
The given data is
No Fusion :: 2748 2700 2655 2822 2511 3149 3257 3213 3220 2753
Yes Fusion :: 3027 3356 3359 3297 3125 2910 2889 2902
The calculated Means are
x1`= 2902.8
X2`= 3108.125
The standard deviations are
s1= 277.265
s2= 205.87
We are using Student's t- test
We formulate null and alternate hypotheses are
H0 : u1 - u2= 0 against Ha: u1 - u 2≠0
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.025, 16) = ± 2.120
Degrees of freedom is calculated df = υ= n1+n2- 2= 10+8-2= 16
Here the difference between the sample means is x`1- x`2
=2902.8-3108.125
= -205.325
The pooled estimate for the common variance σ² is
Sp² = (n1-1) s1² + (n2-1) s2²/ n1+n2-2
= 9( 277.265)² + 7(3108.125)²/ 16
Sp² = 691,882.922+67,623,087.109/16
Sp= 2066.321
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= -205.325/ 2066.321 √1/10+ 1/8
t= -205.325/980.142
t= -0.2095= -0.21
Conclusion
The calculated value of t= -0.21 does not lie in the critical region t= ± 2.120 Therefore we accept our null hypothesis that there's no difference between strength (psi) of liner specimens when a certain fusion process was used and when this process was not used at the 0.05 significance level
