Question

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.

Answer 1

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.