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You measure 49 turtles' weights, and find they have a mean weight of 80 ounces. Assume the population standard deviation is 6.1 ounces. Based on this, construct a 99% confidence interval for the true population mean turtle weight. Round your answers to 2 decimal places.

User Vandale
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1 Answer

22 votes
22 votes

Answer:

The 99% confidence interval for the true population mean turtle weight is between 77.76 and 82.24 ounces.

Explanation:

We have to find our
\alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:


\alpha = (1 - 0.99)/(2) = 0.005

Now, we have to find z in the Z-table as such z has a p-value of
1 - \alpha.

That is z with a p-value of
1 - 0.005 = 0.995, so Z = 2.575.

Now, find the margin of error M as such


M = z(\sigma)/(√(n))

In which
\sigma is the standard deviation of the population and n is the size of the sample.


M = 2.575(6.1)/(√(49)) = 2.24

The lower end of the interval is the sample mean subtracted by M. So it is 80 - 2.24 = 77.76 ounces.

The upper end of the interval is the sample mean added to M. So it is 80 + 2.24 = 82.24 ounces.

The 99% confidence interval for the true population mean turtle weight is between 77.76 and 82.24 ounces.

User TheESJ
by
3.3k points
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