Please help ill give brainiest
1 answer:
You might be interested in
First we will evaluate: ( substitution: u = x² - y², du = - 2 y dy )
=
( than plug in x and 0 )
=
=
= 1/3 x³ ( then another integration )
= 1/3 * 1/4 = 1/12
=
=
= 1/3 x³ ( then another integration )
= 1/3 * 1/4 = 1/12
Answer:
The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.
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 , and s is the standard error.
54% of US adult Twitter users get at least some news on Twitter.
This means that
The standard error for this estimate was 2.4%
This means that
99% confidence level
So , z is the value of Z that has a pvalue of
, so
The lower bound is:
The upper bound is:
The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.