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(Solved): PARTICIPATION ACTIVITY 6.2.2: Confidence intervals for a population mean with known \sigma . The wei ...



PARTICIPATION ACTIVITY 6.2.2: Confidence intervals for a population mean with known

\sigma

. The weights of 5 squash (in pounds) are

10,17,17.5,18.5

, and 19.5 . The sample weights have a mean of

\bar{x} =16.5

. The accepted population standard deviation for this type of squash is

\sigma =1.25

. What is the sample mean's margin of error at the

90%

confidence level? Type as #.###

Show answer What is the sample mean's margin of error at the

99%

confidence level? Type as: #.###

Show answer Correct 0.920 At the

90%

confidence level, the multiplier is

z^(**)=1.645

Thus, the margin of error is

m=z^(**)(\sigma )/(\sqrt(n)) =1.645(1.25)/(\sqrt(5)) ~~0.920

Correct 1.440 At the

99%

confidence level, the multipler is

z^(**)=2.576

Thus, the margin of error is Farnings upcoming



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