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:
Step-by-step explanation:
<u>Seven less than the product of a number n and 1/6 is no more than 94, translating:</u>
<u>Solving for n:</u>
- n/6 ≤ 94 + 7
- n/6 ≤ 101
- n ≤ 101*6
- n ≤ 606
Answer:
a) No Solution
Step-by-step explanation:
36 - 7p = -7(p - 5)
36 - 7p = -7p + 35
1 - 7p = -7p
1 = 0p
so a) No Solution
Answer:
A. 4 units
Step-by-step explanation:
The answer should be X = Z / 6piy