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:
Yes
Step-by-step explanation:
Answer:
Beth can make 9 2 1/3 foot steps
Step-by-step explanation:
21/2.3=9.1
round to 9
Answer:
Step-by-step explanation:
Answer:
x=5 and y=-2
Step-by-step explanation:
3x+5y=5
-(7x+5y=25)
-4x=-20
x=5
plug 5 into one of the equations. 3(5)+5y=5
y= -2
