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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. = ~
where, = average gold thickness of control-immersion tip plating = 1.5 μm
= average gold thickness of total immersion plating = 1.0 μm
= sample standard deviation of control-immersion tip plating = 0.25 μm
= sample standard deviation of total immersion plating = 0.15 μm
= sample of printed circuit edge connectors plated with control-immersion tip plating = 7
= sample of connectors plated with total immersion plating = 5
Also, =
= 0.216
<em>Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>
So, 99% confidence interval for the difference between the mean population mean, () is ;
P(-3.169 < < 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 < < 3.169) = 0.99
P( <
<
) = 0.99
P( < (
) <
) = 0.99
<u>99% confidence interval for</u> () =
[ ,
]
= [ ,
]
= [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].