In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Eighteen randomly selected plots of land were treated with fertilizer A. The average yield, in pounds, was 457 with a standard deviation of 38. Twelve randomly selected plots were treated with fertilizer B. The average yield was 394 pounds with a standard deviation of 23. Find a 99% confidence interval for the difference between the mean yields for the two fertilizers. (Round down the degrees of freedom to the nearest integer and round the final answers to two decimal places.)

(31.25 ; 94.75)

Step-by-step explanation:

Given :

Sample size, n1 = 18

Mean, x1 = 457

Standard deviation, s1 = 38

Sample size, n2 = 12

Mean, x2 = 394

Standard deviation, s2 = 23

Confidence interval :

x1 - x2 ± Tcritical * Standard Error

x1 - x2 = 457 - 394 = 63

Standard Error = sqrt[(s1²/n1 + s2²/n2)]

Standard Error = sqrt((38^2/18)+ (23^2/12)) =11.149

Tcritical = Tα/2, df

df = n1 + n2 - 2 = 18 + 12 - 2 = 28

Using the Tcritical value calculator :

Tcritical at 0.01, 28 = 2.763

Confidence interval :

x1 - x2 ± Tcritical * Standard Error

63 ± (2.763*11.149)

63 ± 31.74687

Lower boundary = 63 - 31.74687 = 31.253

Upper boundary = 63 + 31.74687 = 94.747

(31.25; 94.75)

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}