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anonymous

  • one year ago

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 ?

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  1. anonymous
    • one year ago
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    @rational

  2. rational
    • one year ago
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    start by finding the "margin of error"

  3. rational
    • one year ago
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    whats the multipilier(Z*) value for 95% confidence interval ? (look up in ur notes)

  4. anonymous
    • one year ago
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    .95

  5. rational
    • one year ago
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    there will be a multiplier value, look up ur notes once

  6. rational
    • one year ago
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    For 95% confidence interval, the multiplier value is `1.96` : |dw:1432814569497:dw|

  7. rational
    • one year ago
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    therefore margin of error is \[1.96*\frac{0.2}{\sqrt{10}} = ?\]

  8. anonymous
    • one year ago
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    I feel like I messed up

  9. rational
    • one year ago
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    messed up what

  10. anonymous
    • one year ago
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    .0153664 is what I got and that does not seem right to me

  11. rational
    • one year ago
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    work it again use ur calculator

  12. anonymous
    • one year ago
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    calc says .1214314622

  13. rational
    • one year ago
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    looks good, round it to 0.12 maybe

  14. rational
    • one year ago
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    so the margin of error is 0.12

  15. rational
    • one year ago
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    subtract that from mean, what do u get ?

  16. rational
    • one year ago
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    16.35-0.12 = ?

  17. anonymous
    • one year ago
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    16.23

  18. rational
    • one year ago
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    yes thats the lower bound of confidence interval

  19. rational
    • one year ago
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    for upper bound simply add 16.35 + 0.12 = ?

  20. anonymous
    • one year ago
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    16.47

  21. anonymous
    • one year ago
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    16.22-16.47

  22. rational
    • one year ago
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    so the confidence interval is \[\large (16.23,~ 16.47)\]

  23. anonymous
    • one year ago
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    Tyty

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