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.

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Answer:

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

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]

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

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

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575\frac{6.1}{\sqrt{49}} = 2.24[/tex]

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.