Answer:
idk. Did you figure it out?
Step-by-step explanation:
It would only be A 7-6/2=4 4>3
Answer:
The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence level for the proportion of all American youngsters who are seriously overweight is (0.137, 0.163)
Ann has 8 more socks then tom
Answer:
Yes it is! good job
Step-by-step explanation: