Question

A sample of 66 obese adults was put on a low-carb diet for a year. The average weight loss was 11 pounds and the standard deviation was 19 pounds. Calculate a 99% lower confidence bound for the true average weight loss. What does the bound say about the confidence that the mean weight loss (in the appropriate population) is positive?

Answer 1

`11-2.385(19)/(√(66))=5.422`

So the one side upper confidence interval is (5.422, infty)

And we can conclude that the minimum value that we expected for the true mean of the weight loss at 99% of confidence is 5.422

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=11` represent the sample mean

`\mu` population mean (variable of interest)

`s=19` represent the population standard deviation

n=66 represent the sample size

Solution to the problem

The degrees of freedom for this case are:

`df = n-1= 66-1= 65`

Since the confidence is 0.99 or 99%, the value of
`\alpha=1-0.99=0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,65)".And we see that
`t_(\alpha/2)=2.385`

Now we have everything in order to replace into formula (1):

`11-2.385(19)/(√(66))=5.422`

So the one side upper confidence interval is (5.422, infty)

And we can conclude that the minimum value that we expected for the true mean of the weight loss at 99% of confidence is 5.422

Answer 2

The 99% lower confidence bound for the true average weight loss is 3.98 pounds.

This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.

Explanation:

We have that 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 Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue 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*(19)/(√(66)) = 6.02`

The lower end of the interval is the sample mean subtracted by M. So it is 11 - 6.02 = 3.98 pounds

The 99% lower confidence bound for the true average weight loss is 3.98 pounds.

This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.