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?

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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

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

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.

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option