Question

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 av- erage gold thickness was 1.0 μm, with a standard deviation of 0.15 μm. Find a 99% confidence interval for the difference between the mean thicknesses produced by the two methods.

Answer 1

The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.

( 0.17971 , 0.82028)

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
`se(x^(-) _(1)-x^(-) _(2) ) = \sqrt{(s^2_(1) )/(n_(1) ) +(s^2_(2) )/(n_(2) ) }`

`se(x^(-) _(1)-x^(-) _(2) ) = \sqrt{(0.25^2 )/(7 ) +(0.15^2 )/(5 ) } = 0.115879`

[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)

Answer 2

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

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. =
`\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` ~
`t__n__1+_n__2-2`

where,
`\bar X_1` = average gold thickness of control-immersion tip plating = 1.5 μm

`\bar X_2` = average gold thickness of total immersion plating = 1.0 μm

`s_1` = sample standard deviation of control-immersion tip plating = 0.25 μm

`s_2` = sample standard deviation of total immersion plating = 0.15 μm

`n_1` = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

`n_2` = sample of connectors plated with total immersion plating = 5

Also,
`s_p=\sqrt{((n_1-1)s_1^(2)+(n_2-1)s_2^(2) )/(n_1+n_2-2) }` =
`\sqrt{((7-1)* 0.25^(2)+(5-1)* 0.15^(2) )/(7+5-2) }` = 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, (
`\mu_1-\mu_2`) is ;

P(-3.169 <
`t_1_0` < 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 <
`\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` < 3.169) = 0.99

P(
`-3.169 * {s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` <
`{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}` <
`3.169 * {s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` ) = 0.99

P(
`(\bar X_1-\bar X_2)-3.169 * {s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` < (
`\mu_1-\mu_2`) <
`(\bar X_1-\bar X_2)+3.169 * {s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` ) = 0.99

99% confidence interval for (
`\mu_1-\mu_2`) =

[
`(\bar X_1-\bar X_2)-3.169 * {s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` ,
`(\bar X_1-\bar X_2)+3.169 * {s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` ]

= [
`(1.5-1.0)-3.169 * {0.216\sqrt{(1)/(7)+(1)/(5) } }` ,
`(1.5-1.0)+3.169 * {0.216\sqrt{(1)/(7)+(1)/(5) } }` ]

= [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].