Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately norm
1 answer:
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Complete the square for x and y.
7x^2 + 3y^2 - 28x - 12y = -19
7x^2 - 28x + 3y^2 - 12y = -19
7(x^2 - 4x) + 3(y^2 - 4y) = -19
7(x^2 - 4x + 4) + 3(y^2 - 4y + 4) = -19 + 28 + 12
7(x - 2)^2 + 3(y - 2)^2 = 21
7x^2 + 3y^2 - 28x - 12y = -19
7x^2 - 28x + 3y^2 - 12y = -19
7(x^2 - 4x) + 3(y^2 - 4y) = -19
7(x^2 - 4x + 4) + 3(y^2 - 4y + 4) = -19 + 28 + 12
7(x - 2)^2 + 3(y - 2)^2 = 21
Answer:
The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).
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 z-score that has a p-value of .
Out of 100 people sampled, 42 had kids.
This means that
99% confidence level
So , z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).