For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
Answer:
x = 4
Explanation:
Given the expression;
![\sqrt[]{x}-4\text{ = -2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7Bx%7D-4%5Ctext%7B%20%3D%20-2%7D)
Add 4 to both sides
![\begin{gathered} \sqrt[]{x}-4+4\text{ = -2+4} \\ \sqrt[]{x}=\text{ 2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csqrt%5B%5D%7Bx%7D-4%2B4%5Ctext%7B%20%3D%20-2%2B4%7D%20%5C%5C%20%5Csqrt%5B%5D%7Bx%7D%3D%5Ctext%7B%202%7D%20%5Cend%7Bgathered%7D)
Square both sides
![\begin{gathered} (\sqrt[]{x})^2=2^2 \\ x\text{ = 4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28%5Csqrt%5B%5D%7Bx%7D%29%5E2%3D2%5E2%20%5C%5C%20x%5Ctext%7B%20%3D%204%7D%20%5Cend%7Bgathered%7D)
Hence the value of x is 4
<span>c. 4.6
21 X .22= 4.6
Calculating the variance requires finding the product of 21 and 22%. To make this easier we convert 22% into it's decimal form and construct the equation. To back check this answer we can use 10% of 21 voters which equals 2.1% then double that amount to reach 4.2%, knowing that we now have a close approximation of the variance we can eliminate answers a, b, and d, leaving c as the only logical choice.</span>
The given equation is 4 - 8 * [3/4 + 1/4]
- 4 * [3/4 + 1/4]
- 4 * 1 = - 4
The answer is - 4.
