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:
the answer i got was a=−8b+24/b
Step-by-step explanation:
hoped I helped:)
1350$ is the right answer to your problem trust me I know this
Answer:
10=7
Step-by-step explanation:
6+4=10
2x5=10
10-3=7
Hope this helps :)
-1 + 10 = 9 +10 = 19 + 10 = 29 so the next would be 39 because you add ten to it and there is only one answer with 39 so it's C
