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% con

Question

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

Accepted Answer

Expert Answer to - PARTICIPATION ACTIVITY 6.2.2: Confidence intervals for a population mean with known  sigma . The wei

Answer

Solution for - PARTICIPATION ACTIVITY 6.2.2: Confidence intervals for a population mean with known  sigma . The wei

Answer

This an additional answer to - PARTICIPATION ACTIVITY 6.2.2: Confidence intervals for a population mean with known  sigma . The wei

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