We have a sample of 50 micro-drills for drilling holes in low-carbon alloy steel. The average lifetime, expressed as the number of holes drilled before failure, is 12.68. The standard deviation of the sample of 50 is 6.83 holes. Find the 95% confidence interval for this sample.
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Expert's answer
2020-05-04T19:18:22-0400
The Central Limit Theorem
Let X1,X2,...,Xn be a random sample from a distribution with mean μ and variance σ2. Then if n is sufficiently large, Xˉ has approximately a normal distribution with μXˉ=μ and σXˉ2=σ2/n.
If n>30, the Central Limit Theorem can be used.
We need to construct the 95% confidence interval for the population mean μ. The following information is provided:
Xˉ=12.68,σ=6.83,n=50,
The critical value for α=0.05 is zc=z1−α/2=1.96. The corresponding confidence interval is computed as shown below:
CI=(Xˉ−zcnσ,Xˉ+zcnσ)=
=(12.68−1.96×506.83,12.68+1.96×506.83)≈
≈(10.787,14.573)
Therefore, based on the data provided, the 95% confidence interval for the population mean is 10.787<μ<14.573, which indicates that we are 95% confident that the true population mean μ
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