Question

Fertilizer: In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Ten randomly selected plots of land were treated with fertilizer A, and 9 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
403	466	445	464	437
516	417	420	400	506

Fertilizer B
408	398	382	368	393
437	395	373	424

Construct a  

80%

confidence interval for the difference between the mean yields for the two types of fertilizer. Let

μ1

denote the mean yield for fertilizer A. Use the TI-84 Plus calculator. Round the answers to one decimal place.

The 80% confidence interval for the difference between the mean yields for the two types of fertilizer is <μ1-μ2< .

Answer 1

We are given the harvest of fruits after applying Fertilizer A and B. Let us denote 1 as Fertilizer A and 2 as Fertilizer B.

We can summarize the data for both the fertilizers as

$n_{1}=10,\bar{x}_{1}=447.4,s_{1}=40.5084$

$n_{2}=9,\bar{x}_{2}=397.556,s_{2}=22.6667$

The 80% confidence interval for the difference in means is given by

$CI=\bar{x}_{1}-\bar{x}_{2}\mp t_{1-\alpha /2}*s*\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

The value of $t_{0.90,17}=1.3334$

and s is given by

2 (n1 - 1) si +(n2 – 1)s ni+n2 - 2

$s^{2}=\frac{(10-1)*40.5084^{2}+(9-1)*22.6667^{2}}{10+9-2}$

$s^{2}=\frac{9*1640.9305+8*513.7793}{17}$

$s^{2}=\frac{14768.3745+4110.2343}{17}$

$s^{2}=\frac{18878.6088}{17}$

$s^{2}=1110.5064$

$s=33.3243$

Mean	447.4	397.5556
stdev	40.50844	22.66667

$CI=447.4-397.5556\mp 1.3334*33.3243*\sqrt{\frac{1}{10}+\frac{1}{9}}$

$CI=49.8444\mp 1.3334*33.3243*\sqrt{0.2111}$

$CI=49.8444\mp 1.3334*33.3243*0.4595$

$CI=49.8444\mp 20.4163$

$CI=(29.4281,70.2607)$

The 80% confidence interval for the difference between the mean yields for the two types of fertilizer is 29.4 <μ1-μ2<70.3