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 #.###
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Show answer What is the sample mean's margin of error at the
99%
confidence level? Type as: #.###
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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