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

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"

Step-by-step 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".

represent the sample mean for the sample  

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:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.99 or 99%, the value of and , 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

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

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"

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step-by-step explanation: