Respuesta :
Answer:
The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance
Null hypothesis is rejected
Alternative hypothesis is accepted
The extra carbonation of cola results in a higher average compression strength
Step-by-step explanation:
Step(i):-
Given data
First sample size (n₁) = 10
mean of the first sample(x₁⁻) = 537
standard deviation of the first sample (S₁) = 22
second sample size (n₂) = 10
mean of the second sample (x₂⁻) = 559
standard deviation of the second sample (S₂) = 17
Step(ii):-
Null hypothesis : H₀:- The extra carbonation of cola results in a lower average compression strength
Alternative Hypothesis :H₁
The extra carbonation of cola results in a higher average compression strength
Step(iii):-
By using student's t -test for difference of means
Test statistic
[tex]t = \frac{x^{-} _{1}-x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } } )}[/tex]
where
[tex]S^{2} = \frac{n_{1}S^{2} _{1} + n_{2} S_{2} ^{2} }{n_{1}+n_{2} -2 }[/tex]
[tex]S^{2} = \frac{10(22)^{2} + 10 (17) ^{2} }{10+10 -2 } = \frac{ 7730}{18} = 429.4[/tex]
[tex]t = \frac{537-559 }{\sqrt{429.4 (\frac{1}{10 }+\frac{1}{10 } } )}[/tex]
t = -2.375
|t| = |-2.375| = 2.375
Degrees of freedom
γ = n₁+n₂ -2 = 10+10-2 =18
[tex]t_{\frac{\alpha }{2} ,n-1}=t_{(\frac{0.05}{2} ,18)} = t_{(0.025,18)}}=2.1009[/tex]
The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance
Null hypothesis is rejected
Alternative hypothesis is accepted
Final answer:-
The extra carbonation of cola results in a higher average compression strength
