Respuesta :
Answer:
a) Calculated value F = 1.088 < 3.29
null hypothesis is accepted
Therefore there is no difference between the variances.
b) t = 8.79 > 1.746 for 16 degrees of freedom at 95% level of significance.
The null hypothesis is rejected.
we do not compare means between the two data sets
c) The 95 % of confidence intervals for the difference between means
are (3.30 ,4.68)
Step-by-step explanation:
Given data
Data set A Data set B
Total 677.98 Total 574.24,
n₁= 10 n₂=8
mean((x₁⁻) = 67.798 mean( x₂⁻) = 71.78
variance(S₁²) = 0.663084 variance(S₂²)= 0.727143
S.D(S₁) = 0.814299972 S.D(S₂) = 0.8527267191
a) Null hypothesis (H₀) : σ₁² = σ₂²
Alternative hypothesis: σ₁² ≠ σ₂²
The compare of variances we will use the test statistic is F - distribution
[tex]F = \frac{S^{2} _{2} }{S^{2} _{1} } ( S^{2} _{2}>S^{2} _{1})[/tex]
given data total= 677.98
∑(x-x⁻ )² = 677.98
Given data ∑(y-y⁻ )² = 574.24
S₁² = [tex]\frac{∑(x-x⁻ )²}{n_{1} -1} = \frac{677.98}{10-1} =75.33[/tex]
[tex]S2² = \frac{∑(y-y⁻ )²}{n_{2} -1} = \frac{574.24}{8-1} =82.03[/tex]
[tex]F = \frac{82.03}{75.33} =1.0889[/tex]
The degrees of freedom ν = ( n₁-1, n₂-1) = (10-1 ,8-1) = (9 , 7)
Tabulated value of F for (9 , 7) degrees of freedom at 5% level of significance is 3.29 (see 'F' table at 0.05 level )
Conclusion:-
Calculated value F = 1.088 < 3.677
null hypothesis is accepted
Therefore there is no difference between the variances.
b)
Null hypothesis (H₀) : μ₁ - μ₂≤ D ( given data D= 0.5 add to each data
Alternative hypothesis: μ₁ - μ₂≥ D
Sample statistic :- (x₁⁻ - x₂⁻)
Estimated standard error:-
standard error (S. e) = [tex]\sqrt{\frac{S_{1} ^{2} }{n_{1} } +\frac{S_{2} ^{2} }{n_{2} } }[/tex]
Test statistic 't'
[tex]t = \frac{|x^{-} _{1}-x^{-} _{2}|_-D }{\sqrt{\frac{S_{1} ^{2} }{n_{1} } +\frac{S_{2} ^{2} }{n_{2} } }}[/tex]
mean((x₁⁻) = 67.798 mean( x₂⁻) = 71.78
variance(S₁²) = 0.663084 variance(S₂²)= 0.727143
S.D(S₁) = 0.814299972 S.D(S₂) = 0.8527267191
standard error (S. e) = [tex]\sqrt{\frac{S_{1} ^{2} }{n_{1} } +\frac{S_{2} ^{2} }{n_{2} } }[/tex] = 0.396
Now the test statistic
t = 8.79
The degrees of freedom of t- distribution is
ν= n₁+ n₂-2 = 10 +8-2=16
tabulated value = 1.746 for 16 degrees of freedom at 95% level of significance.
Conclusion:-
The null hypothesis is rejected.
we do not compare means between the two data sets
C) 95% of confidence interval for the difference between means
Solution:-
|(x₁⁻ - x₂⁻)| ± tₐ S. e ( (x₁⁻ - x₂⁻)
where standard error (S. e) = [tex]\sqrt{\frac{S_{1} ^{2} }{n_{1} } +\frac{S_{2} ^{2} }{n_{2} } }[/tex]
given data
mean((x₁⁻) = 67.798 mean( x₂⁻) = 71.78
variance(S₁²) = 0.663084 variance(S₂²)= 0.727143
S.D(S₁) = 0.814299972 S.D(S₂) = 0.8527267191
Standard error = [tex]\sqrt{\frac{0.663084 }{10} +\frac{0.7271 }{8} }[/tex] = 0.396
Standard error = 0.396
The degrees of freedom of t- distribution is
ν= n₁+ n₂-2 = 10 +8-2=16
tabulated value = 1.746 for 16 degrees of freedom at 95% level of significance.
The confidence intervals are
( |(x₁⁻ - x₂⁻)| - tₐ S. e ( (x₁⁻ - x₂⁻), |(x₁⁻ - x₂⁻)| + tₐ S. e ( (x₁⁻ - x₂⁻))
now substitute all values , we get
(|67.798-71.78|-1.746 X (0.396),(|67.798-71.78|+1.746 X (0.396))
on calculation we get
(3.98-0.691 ,3.98+0.691 )
(3.30 ,4.68)
The 95 % of confidence intervals for the difference between means
are (3.30 ,4.68)
