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

Answer:

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.

Step-by-step explanation:

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

$\alpha = \frac{1-0.99}{2} = 0.005$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575*\frac{19}{\sqrt{66}} = 6.02$

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.

Answer 2

Answer:

$11-2.385\frac{19}{\sqrt{66}}=5.422$  

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

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".  

$\bar X=11$ 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):  

$11-2.385\frac{19}{\sqrt{66}}=5.422$  

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