Respuesta :
Answer:
99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].
Step-by-step explanation:
We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.
Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.
Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;
P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n__1+_n__2-2[/tex]
where, [tex]\bar X_1[/tex] = average gold thickness of control-immersion tip plating = 1.5 μm
[tex]\bar X_2[/tex] = average gold thickness of total immersion plating = 1.0 μm
[tex]s_1[/tex] = sample standard deviation of control-immersion tip plating = 0.25 μm
[tex]s_2[/tex] = sample standard deviation of total immersion plating = 0.15 μm
[tex]n_1[/tex] = sample of printed circuit edge connectors plated with control-immersion tip plating = 7
[tex]n_2[/tex] = sample of connectors plated with total immersion plating = 5
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }[/tex] = 0.216
Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.
So, 99% confidence interval for the difference between the mean population mean, ([tex]\mu_1-\mu_2[/tex]) is ;
P(-3.169 < [tex]t_1_0[/tex] < 3.169) = 0.99 {As the critical value of t at 10 degree of
freedom are -3.169 & 3.169 with P = 0.5%}
P(-3.169 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 3.169) = 0.99
P( [tex]-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99
P( [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99
99% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =
[ [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]
= [ [tex](1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] , [tex](1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] ]
= [0.099 μm , 0.901 μm]
Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].
Answer:
The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.
( 0.17971 , 0.82028)
Step-by-step explanation:
Step:-(i)
Given data the average gold thickness was 1.5 μm, with a standard deviation of 0.25 μ m.
Given sample size n₁ = 7
mean of first sample x₁⁻ =1.5 μ m.
Standard deviation of first sample S₁ = 0.25 μ m
Given data Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μ m, with a standard deviation of 0.15 μ m.
Given second sample size n₂ = 5
The mean of second sample x⁻₂ = 1.0 μ m
Standard deviation of first sample S₂ = 0.15 μ m
Level of significance ∝ =0.01 or 99%
Degrees of freedom γ = n₁+ n₂ -2 = 7+5-2=10
tabulated value t = 2.764
Step(ii):-
The 99% of confidence intervals for μ₁- μ₂ is determined by
(x₁⁻ - x⁻₂) - z₀.₉₉ Se((x₁⁻ - x⁻₂) , (x₁⁻ - x⁻₂) + z₀.₉₉ Se((x₁⁻ - x⁻₂)
where [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }[/tex]
[tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{0.25^2 }{7 } +\frac{0.15^2 }{5 } } = 0.115879[/tex]
[1.5-1.0 - 2.764 (0.115879) , (1.5+1.0) + 2.764(0.115879 ]
(0.5-0.32029,0.5+0.32029
( 0.17971 , 0.82028)
Conclusion:-
The 99% of confidence intervals for μ₁- μ₂ is determined by
( 0.17971 , 0.82028)
