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15 votes
15 votes
You measure 40 watermelons' weights, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 11.4 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight.

User LucaRoverelli
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1 Answer

23 votes
23 votes

Answer:


35.36,44.64

Explanation:

Sample size
n=40

Mean weight
\=x =67

Standard deviation
\sigma=11.4

Confidence Interval
CI=0.99

\alpha==0.01

Therefore


Z_{(\alpha)/(2)}=Z_[0.005]

From table


Z_{(\alpha)/(2)}=2.576

Generally, the equation for Margin of error is mathematically given by


M.E=Z_{(\alpha)/(2)}*(\frac{\sigma}{sqrt{n}}


M.E=2.576*((11.4)/(√(40))


M.E=4.64

Therefore Estimated mean is


\=x-M.E<\mu <\=x +E


40-4.64<\mu< 40+4.64


35.36<\mu < 44.64


35.36,44.64

User Jdpjamesp
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