anonymous
  • anonymous
A lathe machine produces machine parts whose lengths are normally distributed with a standard deviation of 0.20 mm. A sample of 10 parts is found to have a mean of 16.35 mm. Based on this sample, what is the 95% confidence interval for the population mean ?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
rational
  • rational
start by finding the "margin of error"
rational
  • rational
whats the multipilier(Z*) value for 95% confidence interval ? (look up in ur notes)

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anonymous
  • anonymous
.95
rational
  • rational
there will be a multiplier value, look up ur notes once
rational
  • rational
For 95% confidence interval, the multiplier value is `1.96` : |dw:1432814569497:dw|
rational
  • rational
therefore margin of error is \[1.96*\frac{0.2}{\sqrt{10}} = ?\]
anonymous
  • anonymous
I feel like I messed up
rational
  • rational
messed up what
anonymous
  • anonymous
.0153664 is what I got and that does not seem right to me
rational
  • rational
work it again use ur calculator
anonymous
  • anonymous
calc says .1214314622
rational
  • rational
looks good, round it to 0.12 maybe
rational
  • rational
so the margin of error is 0.12
rational
  • rational
subtract that from mean, what do u get ?
rational
  • rational
16.35-0.12 = ?
anonymous
  • anonymous
16.23
rational
  • rational
yes thats the lower bound of confidence interval
rational
  • rational
for upper bound simply add 16.35 + 0.12 = ?
anonymous
  • anonymous
16.47
anonymous
  • anonymous
16.22-16.47
rational
  • rational
so the confidence interval is \[\large (16.23,~ 16.47)\]
anonymous
  • anonymous
Tyty

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