Question
In: Math
A biologist measures the lengths of a random sample 45 mature brown trout in a large...
A biologist measures the lengths of a random sample 45 mature brown trout in a large lake and finds that the sample a mean weight of 41 pounds. Assume the population standard deviation is 3.7 pounds. Based on this, construct a 99% confidence interval for the mean weight of all mature brown trout in the lake. Round your anwers to two decimal places. < μ
Solutions
Expert Solution
Solution :
Given that,
= 41
= 3.7
n = 45
At 99% confidence level the z is ,
= 1 - 99% =
1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 =
2.576
Margin of error = E = Z/2* ( /
= 2.576 * (3.7 /
= 1.42
At 99% confidence interval estimate of the population mean is,
- E < 
< + E
41 - 1.42 < < 41 + 1.42
39.58 < < 42.42
milcah answered 7 months agoRelated Solutions
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