Question

A sample of 66 obese adults was put on a lowcarbohydrate diet for a year. The average weight loss was 11 lb and the standard deviation was 19 lb. Calculate a 99% lower confidence bound for the true average weight loss. What does the bound say about confidence that the mean weight loss is positive

Answer 1

`11-2.385(19)/(√(66))=5.422`

So on this case the one lower 99% confidence interval would be given by (5.422;11)

And the lower bound for this case would be 5.422, and that means "the minimum value for the mean at 99% confidence for the mean"

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X - t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, 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 on this case the one lower 99% confidence interval would be given by (5.422;11)

And the lower bound for this case would be 5.422, and that means "the minimum value for the mean at 99% confidence for the mean"