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
Oh okay I’m not going home I’m sorry I don’t have to sleep tueitu lol I
Account A.....with 250 checks
0.20(250) = 50
Account B...with 250 checks
0.10(250) + 1.50 = 25 + 1.50 = 26.50
so by using Account B, u r saving (50 - 26.50) = 23.50
Answer:
Please see the graph below
Step-by-step explanation:
The table is just selecting numbers for x and then solving for what y would be. Once we have the ordered pairs (x,y) We can graph the points and see the line.
