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: True
Step-by-step explanation: This is because they are equivalent fractions
Hope this helps and have a nice day
Answer:
A
Step-by-step explanation:
refer to the pic above
Answer:
50 + 11.5x = 188
Step-by-step explanation:
Answer:
-39991'
Step-by-step explanation:
We require the sum of 7 terms of the G.S. -1, 6, -36 .....
The common ratio r = 6/-1 = -6 and the first term = -1.
Sum of n terms = a1 . (r^n - 1) / (r - 1)
Sum of 7 terms = -1 * (-6)^7 - 1) / (-6 - 1)
= -1 * 39991
= -39991 (answer).
