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Answer:
The 90% confidence interval for the mean weight of all adult male grizzly bears in the United States is between 573 pounds and 649 pounds.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 23 - 1 = 2
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 22 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 2.0739
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.0739\frac{89}{\sqrt{23}} = 38[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 611 - 38 = 573 pounds
The upper end of the interval is the sample mean added to M. So it is 611 + 38 = 649 pounds
The 90% confidence interval for the mean weight of all adult male grizzly bears in the United States is between 573 pounds and 649 pounds.
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