In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Thirteen randomly selected plots of land were treated with fertilizer A, and 8 randomly selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured from each plot. Following are the results. Fertilizer A
445 510 464 472 441 480 403 460 448
457 437 505 417

Fertilizer B
398 380 368 393 424 387 378 415

Required:
Construct a 95% confidence interval for the difference between the mean yields for the two types of fertilizer.

The 95% confidence interval for the difference between the mean yields for the two types of fertilizer is approximately 92.5536 pounds < < 35.39 pounds

Step-by-step explanation:

The harvest from each plot treated with fertilizer A and fertilizer B are presented as follows;

Fertilizer A

445, 510, 464, 472, 441, 480, 403, 460, 448, 457, 437, 505, 417

Fertilizer B

398, 380, 368 393, 424, 387, 378, 415

The number of trials of fertilizer A, n₁ = 13

The number of trials of fertilizer B, n₂ = 8

The confidence level = 95%

α = 1 - 0.95 = 0.05

∴ α/2 = 0.025

The degrees of freedom, df = 8 - 1 = 7

= 2.365

From Microsoft Excel, we get;

The standard deviation of the pounds of harvested fruits for the plots treated with fertilizer A s₁ = 30.71603

The mean of the pounds of harvested fruits for the plots treated with fertilizer A, = 456.8462

The standard deviation of the pounds of harvested fruits for the plots treated with fertilizer B s = 18.99201

The mean of the pounds of harvested fruits for the plots treated with fertilizer B,   = 392.875

The 95% confidence interval, C.I. for the difference between the mean yields for the two types of fertilizer can be constructed with the following formula

By variable substitution, we get;

The 95% confidence interval for the difference between the mean yields for the two types of fertilizer , which can be expressed as 92.5536 lbs. < < 35.39 lbs.

there still no table

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !